Motion in 1D

To find the position or distance from the origin at $t = 8\text{ s}$, we calculate the net area under the $v-t$ curve up to $t = 8\text{ s}$.
The motion can be divided into two regions: above the time axis ($0\text{ s}$ to $4\text{ s}$) and below the time axis ($4\text{ s}$ to $8\text{ s}$).
Area above the time axis forms a trapezoid from $t = 0\text{ s}$ to $t = 4\text{ s}$ with height $h_1 = 4\text{ m/s}$, top base $b_1 = 3 - 1 = 2\text{ s}$, and bottom base $B_1 = 4 - 0 = 4\text{ s}$.
$\text{Area}_{\text{above}} = \frac{1}{2} \times (B_1 + b_1) \times h_1 = \frac{1}{2} \times (4 + 2) \times 4 = 12\text{ m}$
Area below the time axis forms a trapezoid from $t = 4\text{ s}$ to $t = 8\text{ s}$ with height $h_2 = 2\text{ m/s}$, top base $b_2 = 7 - 5 = 2\text{ s}$, and bottom base $B_2 = 8 - 4 = 4\text{ s}$.
$\text{Area}_{\text{below}} = \frac{1}{2} \times (B_2 + b_2) \times h_2 = \frac{1}{2} \times (4 + 2) \times 2 = 6\text{ m}$
The displacement from the origin at $t = 8\text{ s}$ is:
$\text{Displacement} = \text{Area}_{\text{above}} - \text{Area}_{\text{below}} = 12 - 6 = 6\text{ m}$
Thus, the distance of the particle from the origin after 8 seconds is $6\text{ meters}$.

First, calculate the total distance travelled in 9 seconds, which corresponds to the area of the trapezoid from $t = 1\text{ s}$ to $t = 9\text{ s}$.
The trapezoid has parallel sides of lengths $b_1 = 5 - 3 = 2\text{ s}$ and $b_2 = 9 - 1 = 8\text{ s}$, and height $h = 10\text{ m/s}$.
$\text{Distance in 9 seconds} = \frac{1}{2} \times (8 + 2) \times 10 = 50\text{ m}$
Next, calculate the distance travelled in the last 4 seconds, which is the time interval from $t = 5\text{ s}$ to $t = 9\text{ s}$.
This region forms a triangle with base $b = 9 - 5 = 4\text{ s}$ and height $h = 10\text{ m/s}$.
$\text{Distance in last 4 seconds} = \frac{1}{2} \times 4 \times 10 = 20\text{ m}$
Now, find the ratio of distance travelled in the last 4 seconds to the total distance travelled in 9 seconds:
$\text{Ratio} = \frac{20}{50} = \frac{2}{5}$
Thus, the correct ratio is $\frac{2}{5}$.

To find the displacement and distance, we calculate the area of each rectangular section from $t = 0\text{ s}$ to $t = 6\text{ s}$.
For $t = 0\text{ s}$ to $t = 2\text{ s}$, area $A_1 = 2 \times 4 = 8\text{ m}$.
For $t = 2\text{ s}$ to $t = 4\text{ s}$, area $A_2 = 2 \times (-2) = -4\text{ m}$.
For $t = 4\text{ s}$ to $t = 6\text{ s}$, area $A_3 = 2 \times 2 = 4\text{ m}$.
Now, calculate total displacement:
$\text{Displacement} = A_1 + A_2 + A_3 = 8 - 4 + 4 = 8\text{ m}$
Next, calculate total distance:
$\text{Distance} = \vert{}A_1\vert{} + \vert{}A_2\vert{} + \vert{}A_3\vert{} = 8 + 4 + 4 = 16\text{ m}$
Therefore, the displacement and distance travelled by the body in 6 s are $8\text{ m}$ and $16\text{ m}$, respectively.

To find the distance traversed in 3 seconds, we need to calculate the area under the $v-t$ graph from $t = 0\text{ s}$ to $t = 3\text{ s}$.
The area can be divided into three geometric parts:
1. Triangle from $t = 0\text{ s}$ to $t = 1\text{ s}$ with base $1\text{ s}$ and height $20\text{ m/s}$:
$\text{Area}_1 = \frac{1}{2} \times 1 \times 20 = 10\text{ m}$
2. Rectangle from $t = 1\text{ s}$ to $t = 2\text{ s}$ with base $1\text{ s}$ and height $20\text{ m/s}$:
$\text{Area}_2 = 1 \times 20 = 20\text{ m}$
3. Trapezoid from $t = 2\text{ s}$ to $t = 3\text{ s}$ with parallel heights $20\text{ m/s}$ and $10\text{ m/s}$, and width $1\text{ s}$:
$\text{Area}_3 = \frac{1}{2} \times (20 + 10) \times 1 = 15\text{ m}$
Adding all the areas together gives the total distance traversed in 3 seconds:
$\text{Total Distance} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 = 10 + 20 + 15 = 45\text{ m}$
Therefore, the distance traversed by the particle in 3 seconds is $45\text{ m}$.

From the given velocity-time graph, we need to find the time intervals where acceleration is zero.
Acceleration is zero when the graph is a horizontal line (constant velocity).
Looking at the graph, the velocity is constant at $v = 1$ m/s in two intervals:
1. From $t = 0$ s to $t = 20$ s.
2. From $t = 40$ s to $t = 60$ s.
Calculate the displacement for the first interval ($t = 0$ to $20$ s):
Velocity $v = 1$ m/s.
Time $\Delta t_1 = 20 - 0 = 20$ s.
Displacement $s_1 = v \times \Delta t_1 = 1 \times 20 = 20$ m.
Calculate the displacement for the second interval ($t = 40$ to $60$ s):
Velocity $v = 1$ m/s.
Time $\Delta t_2 = 60 - 40 = 20$ s.
Displacement $s_2 = v \times \Delta t_2 = 1 \times 20 = 20$ m.
Total displacement during the time intervals with zero acceleration is the sum of displacements in these intervals.
Total displacement $S = s_1 + s_2 = 20 \text{ m} + 20 \text{ m} = 40$ m.
This corresponds to option (4).

To find the maximum acceleration, we evaluate the slope ($\frac{\Delta v}{\Delta t}$) for each segment where velocity increases (positive acceleration).
From $t = 0\text{ s}$ to $t = 20\text{ s}$, velocity changes from $0\text{ m/s}$ to $20\text{ m/s}$.
$a_1 = \frac{20 - 0}{20 - 0} = 1\text{ m/s}^2$
From $t = 20\text{ s}$ to $t = 30\text{ s}$, velocity remains constant at $20\text{ m/s}$.
$a_2 = 0\text{ m/s}^2$
From $t = 30\text{ s}$ to $t = 40\text{ s}$, velocity changes from $20\text{ m/s}$ to $60\text{ m/s}$.
$a_3 = \frac{60 - 20}{40 - 30} = \frac{40}{10} = 4\text{ m/s}^2$
Comparing the acceleration values, the maximum magnitude of positive acceleration occurs between $t = 30\text{ s}$ and $t = 40\text{ s}$, which is $4\text{ m/s}^2$.
Therefore, the maximum acceleration is $4\text{ m s}^{-2}$.

From the given $v-t$ graph, identify the velocity of the block at $t = 2\text{ sec}$ and $t = 4\text{ sec}$.
At $t_1 = 2\text{ sec}$, velocity $v_1 = 10\text{ ms}^{-1}$.
At $t_2 = 4\text{ sec}$, velocity $v_2 = 0\text{ ms}^{-1}$.
Now, calculate the change in velocity:
$\Delta v = v_2 - v_1 = 0 - 10 = -10\text{ ms}^{-1}$
Calculate the time interval:
$\Delta t = t_2 - t_1 = 4 - 2 = 2\text{ sec}$
Substitute these values into the average acceleration formula:
$a_{\text{avg}} = \frac{-10}{2} = -5\text{ m/s}^2$
Thus, the average acceleration from $t = 2\text{ sec}$ to $t = 4\text{ sec}$ is $-5\text{ m/s}^2$.

The particle starts from rest, so initial velocity $v_{\text{initial}} = 0\text{ m/s}$.
The maximum speed occurs when the acceleration becomes zero (at $t = 8\text{ s}$), as velocity continues to increase as long as acceleration is positive.
The area under the triangular $a-t$ graph from $t = 0\text{ s}$ to $t = 8\text{ s}$ represents the change in velocity:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Area} = \frac{1}{2} \times 8\text{ s} \times 5\text{ m/s}^2 = 20\text{ m/s}$
Therefore, maximum speed $v_{\text{max}} = v_{\text{initial}} + \text{Area} = 0 + 20 = 20\text{ m/s}$.

The particle starts from rest, so initial velocity $u = 0\text{ m/s}$.
The velocity at $t = 4\text{ s}$ is given by $v = u + \text{Area under } a-t \text{ graph from } t = 0\text{ s to } t = 4\text{ s}$.
The area under the $a-t$ graph consists of two parts:
1. Triangular region from $t = 0\text{ s}$ to $t = 2\text{ s}$:
$\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 10 = 10\text{ m/s}$
2. Rectangular region from $t = 2\text{ s}$ to $t = 4\text{ s}$:
$\text{Area}_2 = \text{width} \times \text{height} = (4 - 2) \times 10 = 2 \times 10 = 20\text{ m/s}$
Total area under the graph is:
$\text{Total Area} = \text{Area}_1 + \text{Area}_2 = 10 + 20 = 30\text{ m/s}$
Therefore, the velocity at $t = 4\text{ sec}$ is:
$v = 0 + 30 = 30\text{ m/s}$

Looking at the $t-s$ graph, the time axis is represented vertically and the displacement axis is represented horizontally.
Between points B and C, as displacement decreases, time is shown moving backwards or having multiple values of time for a given position.
Time is a non-decreasing variable in real physical systems and can never decrease or flow backwards.
Furthermore, a single value of displacement $s$ cannot correspond to multiple values of time in this manner for a single continuous path without representing impossible conditions.
Therefore, such a position-time or time-position curve cannot represent any real motion.
Hence, the graph shown is impossible.

From the given acceleration-time graph, we analyze the motion in two intervals.
Interval 1: $t = 0$ to $t = 2$ s.
The acceleration is constant at $a = 3$ m/s$^2$.
Since $a = \frac{dv}{dt}$, the slope of the velocity-time graph is positive and constant.
Assuming initial velocity $u = 0$ m/s (as suggested by graphs 1 and 3 starting at origin), the velocity at $t = 2$ s is $v = u + at = 0 + 3(2) = 6$ m/s.
The graph should be a straight line from $(0, 0)$ to $(2, 6)$.
Interval 2: $t = 2$ to $t = 4$ s.
The acceleration is constant at $a = -3$ m/s$^2$.
The slope of the velocity-time graph is negative and constant.
The initial velocity for this interval is the final velocity of the previous interval, $v(2) = 6$ m/s.
The velocity at $t = 4$ s is $v = v(2) + a(t - 2) = 6 + (-3)(4 - 2) = 6 - 6 = 0$ m/s.
The graph should be a straight line from $(2, 6)$ to $(4, 0)$.
Combining both intervals, the velocity-time graph starts at 0, increases linearly to 6 m/s at $t=2$ s, and then decreases linearly back to 0 m/s at $t=4$ s.
This shape corresponds to Graph (1).
Therefore, the correct option is (1), which corresponds to A.

We analyze the motion in three time intervals based on the given acceleration-time graph. The body starts from rest, so initial velocity $u = 0$ m/s.
Interval 1: $t = 0$ to $t = 2$ s.
The acceleration is constant at $a = 5$ m/s$^2$.
Using $v = u + at$, we get $v = 0 + 5t = 5t$.
At $t = 2$ s, velocity $v = 5(2) = 10$ m/s.
The velocity-time graph is a straight line with a positive slope starting from the origin $(0,0)$ and ending at $(2, 10)$.
Interval 2: $t = 2$ to $t = 6$ s.
The acceleration is $a = 0$ m/s$^2$.
This means the velocity is constant.
So, $v = 10$ m/s throughout this interval.
The velocity-time graph is a horizontal line at $v = 10$ from $t = 2$ to $t = 6$.
Interval 3: $t = 6$ to $t = 10$ s.
The acceleration is constant at $a = -5$ m/s$^2$.
The initial velocity for this interval is the final velocity of the previous interval, which is $10$ m/s.
Using $v = u + a(t - t_0)$, where $u = 10$ and $t_0 = 6$, we get $v = 10 - 5(t - 6)$.
At $t = 10$ s, velocity $v = 10 - 5(10 - 6) = 10 - 20 = -10$ m/s.
The velocity-time graph is a straight line with a negative slope starting from $(6, 10)$ and ending at $(10, -10)$. It crosses the time axis ($v=0$) at $t = 8$ s.
Comparing this behavior with the given options, Graph (1) matches perfectly: it shows a linear increase to 10 m/s, a constant phase at 10 m/s, and then a linear decrease to -10 m/s.
Therefore, the correct option is Graph (1), which corresponds to A.

From the given $v-x$ graph, we can determine the equation of the line.
The graph is a straight line with y-intercept $v_0 = 30$ and x-intercept $x_0 = 10$ (assuming the value is 10 based on visual inspection, though the exact value doesn't change the shape, let's verify with the options). Wait, looking at the options, the intercept is -90. Let's derive the general form.
Let the equation of the line be $v = -mx + c$, where $m$ is the magnitude of the slope and $c$ is the y-intercept ($30$).
The slope of the $v-x$ graph is $\frac{dv}{dx} = -m$.
Using the formula $a = v \frac{dv}{dx}$, we substitute $v$ and $\frac{dv}{dx}$:
$a = (-mx + c)(-m) = m^2x - mc$.
This equation shows that acceleration $a$ varies linearly with displacement $x$.
The graph of $a$ vs $x$ should be a straight line with a positive slope ($m^2$) and a negative y-intercept ($-mc$).
From the given graph, $c = 30$. Let's assume the x-intercept is $10$. Then slope $m = \frac{30-0}{10-0} = 3$.
So, $a = (3)^2 x - (3)(30) = 9x - 90$.
At $x = 0$, $a = -90$.
At $x = 10$, $a = 0$.
The graph should start at $a = -90$ on the vertical axis and increase linearly with a positive slope, crossing the x-axis at a positive value.
Looking at the options:
Graph (1) shows a line starting at $-90$ on the a-axis and going upwards with a positive slope. This matches our derivation.
Graph (2) has a negative slope.
Graph (3) is a constant negative value (zero slope).
Graph (4) has a negative slope starting from -90.
Therefore, Graph (1) is the correct representation.

Analyze the motion from the given $x-t$ graph:
From $t = 0$ to $t = T$: The slope of the $x-t$ graph is positive but decreasing. This indicates the particle is moving in the positive direction with decreasing velocity (decelerating). The speed is decreasing.
At $t = T$: The slope is zero (tangent is horizontal). The velocity is zero, so the particle is momentarily at rest.
From $t = T$ to $t = 2T$: The slope is negative and its magnitude is increasing (the curve becomes steeper downwards). This indicates the particle is moving in the negative direction with increasing speed (accelerating in reverse).
Analyze the distance-time graph:
Distance is the total path length, so it must always increase (or remain constant). It cannot decrease. This eliminates graphs (2) and (3) which show the value going down.
The slope of the distance-time graph corresponds to the speed ($|v|$).
From $t = 0$ to $t = T$: Speed is decreasing. So, the slope of the distance graph should decrease. The curve should be concave down (bending downwards).
At $t = T$: Speed is zero. So, the slope of the distance graph should be zero (horizontal tangent).
From $t = T$ to $t = 2T$: Speed is increasing. So, the slope of the distance graph should increase. The curve should be concave up (bending upwards).
Comparing with the options:
Graph (1) shows a curve that is always increasing. It starts with a steep slope, the slope decreases (concave down) until it flattens around $t=T$, and then the slope increases again (concave up) becoming steep at $t=2T$. This matches the behavior of the speed derived from the $x-t$ graph.
Graph (4) shows a curve where the slope decreases throughout, implying the particle slows down and stops, which is incorrect for the second half of the motion.
Therefore, Graph (1) is the correct representation.

From the equation of motion for constant acceleration starting from rest:
$v^2 = 2as$
This can be written as $v = \sqrt{2a} \cdot \sqrt{s}$.
This equation shows that velocity $v$ is proportional to the square root of displacement $s$ ($v \propto \sqrt{s}$).
The graph of $v$ versus $s$ will be a parabola with its axis along the displacement axis (horizontal axis).
Alternatively, we can look at the slope of the $v-s$ graph. The slope is $\frac{dv}{ds}$.
Differentiating $v^2 = 2as$ with respect to $s$:
$2v \frac{dv}{ds} = 2a$
$\frac{dv}{ds} = \frac{a}{v}$
Since acceleration $a$ is constant and positive, and velocity $v$ increases as displacement $s$ increases, the term $\frac{a}{v}$ decreases.
This means the slope of the $v-s$ graph decreases as $s$ increases.
- Graph (1) shows a constant slope (straight line), which implies $v \propto s$. This is incorrect.
- Graph (2) shows a curve with a decreasing slope (concave down), which matches the relationship $v \propto \sqrt{s}$. This is correct.
- Graph (3) shows a straight line with a non-zero intercept, implying initial velocity is not zero. This is incorrect.
- Graph (4) shows a curve with an increasing slope (concave up), which implies $v \propto s^n$ where $n > 1$. This is incorrect.
Therefore, graph (2) correctly represents the velocity-displacement curve.

Analyze the slope of the $x-t$ graph in each interval:
1. **Interval OA:** The curve is concave up (bending upwards). The slope (velocity) is increasing from the origin. Since velocity is increasing, acceleration is positive ($+$).
2. **Interval AB:** The graph appears to be a straight line with a small positive slope. A straight line on an $x-t$ graph indicates constant velocity. Therefore, acceleration is zero ($0$).
3. **Interval BC:** The curve appears to flatten out or bend downwards (concave down). The slope (velocity) is decreasing. Since velocity is decreasing, acceleration is negative ($-$).
4. **Interval CD:** The graph is a straight line with a steep positive slope. This indicates a constant high velocity. Therefore, acceleration is zero ($0$).
Matching this sequence (+, 0, -, 0) with the given options:
- Option (1): +, 0, +, +
- Option (2): -, 0, +, 0
- Option (3): +, 0, -, 0
- Option (4): -, 0, -, 0
The correct sequence corresponds to Option (3).

When a body is projected vertically upwards, the only acceleration acting on it is the acceleration due to gravity.
This acceleration is constant in magnitude ($9.8$ m/s$^2$) and always acts downwards.
Mathematically, $a = -g$ (taking upward direction as positive).
Since acceleration does not change with time, its value remains constant throughout the upward and downward motion.
Plotting a constant acceleration against time on a graph yields a horizontal straight line parallel to the time axis.
Therefore, the acceleration-time curve is a straight line.

The relationship between displacement $s$ and time $t$ for a body projected vertically upward is $s = ut - \frac{1}{2}gt^2$.
This is a quadratic equation, so the graph between $s$ and $t$ must be a parabola.
Since the term with $t^2$ has a negative coefficient, the parabola opens downwards (concave down).
At $t=0$, $s=0$, so the graph starts from the origin.
The body goes up, reaches a maximum height (vertex of the parabola), and then comes down, so $s$ decreases back to 0.
Graph (1) shows straight lines, implying constant velocity, which is incorrect.
Graph (2) and (3) do not represent the motion starting from zero displacement and returning.
Graph (4) shows a parabolic curve starting from the origin, reaching a maximum, and returning to the axis, which correctly represents the motion.
Therefore, the correct option is (4).

From the velocity-time graph, the rocket reaches maximum height when velocity becomes zero at t = 120s (point B).
The maximum height is equal to the area under the velocity-time graph from t = 0 to t = 120s.
This area consists of two triangles:
Triangle 1 (O to A): From t = 0 to t = 20s with maximum velocity 1000 m/s
Area 1 = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 1000 = 10,000$ m
Triangle 2 (A to B): From t = 20s to t = 120s (duration = 100s) with initial velocity 1000 m/s
Area 2 = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 100 \times 1000 = 50,000$ m
Total height = Area 1 + Area 2 = 10,000 + 50,000 = 60,000 m
Converting to km: 60,000 m = 60 km
Therefore, the maximum height attained by the rocket is 60 km.

From the velocity-time graph, the velocity of the rocket increases from $t = 0$ to $t = 20$ s.
After $t = 20$ s, the velocity starts decreasing, which indicates that retardation begins at this point.
Therefore, the height covered by the rocket before retardation is the displacement from $t = 0$ to $t = 20$ s.
This displacement is equal to the area under the graph for this time interval, which forms a triangle.
Base of the triangle = $20$ s.
Height of the triangle = $1000$ m/s.
Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 1000 = 10,000$ m.
Converting meters to kilometers: $10,000 \text{ m} = 10 \text{ km}$.
This matches option (2).

From the velocity-time graph, the rocket attains maximum height when its velocity becomes zero.
Looking at the graph, velocity becomes zero at $t = 120$ s.
The total time taken to attain maximum height is $\Delta t = 120$ s.
The total displacement (maximum height) is the area under the velocity-time graph from $t = 0$ to $t = 120$ s.
The area of the triangle formed is $\frac{1}{2} \times \text{base} \times \text{height}$.
Base = $120$ s, Height = $1000$ m/s.
Total displacement $\Delta x = \frac{1}{2} \times 120 \times 1000 = 60,000$ m.
Mean velocity $v_{avg} = \frac{\Delta x}{\Delta t} = \frac{60,000}{120}$.
$v_{avg} = 500$ m/s.
This corresponds to option (3).

From the velocity-time graph of Q. no. 152, the rocket undergoes retardation during the segment AB from time $t_1 = 20\text{ s}$ to $t_2 = 120\text{ s}$.
At $t_1 = 20\text{ s}$, the initial velocity is $v_1 = 1000\text{ ms}^{-1}$.
At $t_2 = 120\text{ s}$, the final velocity before it stops climbing (point B) is $v_2 = 0\text{ ms}^{-1}$.
The time interval for retardation is:
$\Delta t = 120 - 20 = 100\text{ s}$
Using the retardation formula:
$\text{Retardation} = \frac{1000 - 0}{100} = 10\text{ m/s}^2$
Therefore, the retardation of the rocket is $10\text{ m/s}^2$.

From the velocity-time graph of Q. no. 152, the rocket accelerates along segment OA from $t_1 = 0\text{ s}$ to $t_2 = 20\text{ s}$.
At $t_1 = 0\text{ s}$, the initial velocity is $v_1 = 0\text{ ms}^{-1}$.
At $t_2 = 20\text{ s}$, the final velocity is $v_2 = 1000\text{ ms}^{-1}$.
The time interval for acceleration is:
$\Delta t = 20 - 0 = 20\text{ s}$
Using the acceleration formula:
$a = \frac{1000 - 0}{20} = 50\text{ m/s}^2$
Therefore, the acceleration of the rocket is $50\text{ m/s}^2$.

From the velocity-time graph of Q. no. 152, let us observe the sign of velocity during different parts of the motion.
Along segment OA, the velocity increases from $0$ to $+1000\text{ m/s}$ ($v > 0$), so the rocket is moving upward with acceleration.
Along segment AB, the velocity decreases from $+1000\text{ m/s}$ to $0\text{ m/s}$ ($v > 0$), so the rocket is still moving upward, but with deceleration/retardation.
Therefore, during the entire path OAB, velocity is positive ($v > 0$), which means the rocket goes up during OAB.
At point B ($t = 120\text{ s}$), velocity becomes zero and then turns negative along segment BC ($v < 0$).
Negative velocity signifies that the rocket has reversed its direction and is coming down.
Thus, the rocket goes up and comes down on parts OAB and BC respectively.

Initial velocity of the ball at $t = 0$ is zero ($u = 0$).
Since downward direction is taken as positive, acceleration $a = +g$.
As the ball falls, its velocity increases linearly with time following $v = gt$, giving a straight line with a positive slope until time $T/2$.
When the ball hits the glass surface, it collides elastically, reversing its velocity direction instantaneously from $+v$ to $-v$.
As it moves upward against gravity, acceleration is still $+g$ downward, so velocity linearly increases from $-v$ to $0$ at time $T$.
This produces a periodic sawtooth waveform with positive slope throughout each falling and rising phase, jumping from positive to negative velocity at collision points.
Comparing with the options shown in the image, Graph (3) accurately represents this motion.

Initially, the ball is dropped from $h = d$, so initial velocity at $h = d$ is $v = 0$.
As the ball falls downward, height $h$ decreases from $d$ to $0$, and velocity $v$ becomes increasingly negative, following the parabolic relation $v = -\sqrt{2g(d - h)}$.
At $h = 0$ (ground), velocity reaches its maximum negative magnitude.
Upon impact, the ball rebounds upward, so velocity abruptly switches sign to positive at $h = 0$.
As the ball moves upward from $h = 0$ to $h = d/2$, height $h$ increases while positive velocity decreases parabolically following $v = +\sqrt{2g(d/2 - h)}$ until $v = 0$ at $h = d/2$.
Comparing with the options shown in the image, Graph (1) correctly represents these parabolic segments in the negative velocity region (downward motion from $d$ to $0$) and positive velocity region (upward motion from $0$ to $d/2$).

The position-time equation with origin at the base of the tower ($x_0 = 60\text{ m}$) is:
$x(t) = 60 + 20t - 5t^2$
At $t = 0\text{ s}$, the initial position is $x(0) = 60\text{ m}$.
The maximum height (peak of the parabola) occurs when velocity becomes zero:
$v(t) = \frac{dx}{dt} = 20 - 10t = 0 \implies t = 2\text{ s}$
The position at maximum height ($t = 2\text{ s}$) is:
$x(2) = 60 + 20(2) - 5(2)^2 = 60 + 40 - 20 = 80\text{ m}$
The stone reaches the ground when position $x(t) = 0$:
$0 = 60 + 20t - 5t^2 \implies t^2 - 4t - 12 = 0 \implies (t - 6)(t + 2) = 0 \implies t = 6\text{ s}$
Therefore, the position-time graph starts at $x = 60\text{ m}$ at $t = 0\text{ s}$, rises to a maximum position of $x = 80\text{ m}$ at $t = 2\text{ s}$, and decreases back to $x = 0\text{ m}$ at $t = 6\text{ s}$.
Comparing this with the given choices in the image, Graph (3) correctly starts at $x = 60\text{ m}$, goes up, and reaches $x = 0\text{ m}$ at $t = 6\text{ s}$.

During the first 4 seconds, the car accelerates uniformly from rest.
The equation of motion is $x = \frac{1}{2}at^2$, so the $x-t$ graph is a parabola curving upwards (slope increases with time).
After 4 seconds, the car moves with a constant/uniform velocity.
The equation of motion becomes $x = x_4 + v(t - 4)$, so the $x-t$ graph becomes a straight line with a constant non-zero slope.
Looking at the choices given, Graph (4) shows a parabolic curve bending upward for the first interval, followed by a straight inclined line for the uniform velocity phase.
Therefore, Graph (4) correctly represents the motion of the car.

From the given $x-t$ graph, at $t = 0\text{ s}$, the initial position of the particle is $x(0) = 10\text{ m}$.
Average velocity over the time interval from $0$ to $t$ is defined as $v_{\text{avg}} = \frac{x(t) - x(0)}{t}$.
For $v_{\text{avg}} = 0$, we require $x(t) - x(0) = 0$, which means $x(t) = x(0) = 10\text{ m}$.
Looking at the graph, the curve passes through $x = 10\text{ m}$ again at $t = 12\text{ sec.}$.
Therefore, at $t = 12\text{ sec.}$, the displacement is zero, making the average velocity during the period $0$ to $12\text{ sec.}$ equal to zero.

From the given $v^2-s$ graph, at initial position $s_1 = 0\text{ km}$, $v_1^2 = 4600\text{ km}^2/\text{hr}^2$.
At position $s_2 = 0.6\text{ km}$, $v_2^2 = 900\text{ km}^2/\text{hr}^2$.
Change in $v^2$ is $\Delta (v^2) = 900 - 4600 = -3700\text{ km}^2/\text{hr}^2$.
Change in position is $\Delta s = 0.6 - 0 = 0.6\text{ km}$.
Using the formula $2a = \frac{\Delta (v^2)}{\Delta s}$:
$2a = \frac{-3700}{0.6} = -6166.67\text{ km/hr}^2$
Acceleration $a = \frac{-6166.67}{2} = -3083.33\text{ km/hr}^2 \approx -3084\text{ km/hr}^2$.
Therefore, the acceleration of the particle is $-3084\text{ km/hr}^2$.

Total time of journey $T = 4\text{ min} = 4 \times 60 = 240\text{ sec}$.
Total distance between stations $S = 2.25\text{ km} = 2250\text{ m}$.
Time of acceleration $t_1 = 40\text{ sec}$.
Time of deceleration $t_3 = 20\text{ sec}$.
Time spent moving at constant maximum speed $v_{\text{max}}$:
$t_2 = T - t_1 - t_3 = 240 - 40 - 20 = 180\text{ sec}$.
The velocity-time graph forms a trapezoid where total distance $S$ is the area under the graph:
$S = \frac{1}{2} t_1 v_{\text{max}} + t_2 v_{\text{max}} + \frac{1}{2} t_3 v_{\text{max}}$
$S = v_{\text{max}} \left( \frac{t_1}{2} + t_2 + \frac{t_3}{2} \right)$
Substitute the given values:
$2250 = v_{\text{max}} \left( \frac{40}{2} + 180 + \frac{20}{2} \right)$
$2250 = v_{\text{max}} (20 + 180 + 10) = 210 v_{\text{max}}$
$v_{\text{max}} = \frac{2250}{210} = \frac{75}{7} \approx 10.71\text{ m/sec}$
Therefore, the maximum speed of the train is $10.7\text{ m/sec}$.

Given acceleration $\alpha = 2\text{ m/s}^2$.
Given retardation $\beta = 4\text{ m/s}^2$.
Total time of motion $T = 3\text{ seconds}$.
Using the direct formula for maximum speed in acceleration-deceleration motion:
$v_{\text{max}} = \frac{\alpha \beta}{\alpha + \beta} T$
Substitute the given values into the formula:
$v_{\text{max}} = \frac{2 \times 4}{2 + 4} \times 3$
$v_{\text{max}} = \frac{8}{6} \times 3 = \frac{24}{6} = 4\text{ m/s}$
Therefore, the maximum speed attained by the car is $4\text{ m/s}$.

$H_{max} = \frac{u^2}{2g} = \frac{(15)^2}{2 \times 10} = 11.25\text{ m}$
Step Solution:
Step 1: Identify the initial parameters given in the problem: initial velocity $u = 15\text{ m/s}$ and acceleration due to gravity $g = 10\text{ m/s}^2$.
Step 2: Use the formula for maximum height in motion under gravity, $H_{max} = \frac{u^2}{2g}$.
Step 3: Substitute the initial velocity $u = 15$ and acceleration due to gravity $g = 10$ into the equation to get $H_{max} = \frac{15^2}{2 \times 10}$.
Step 4: Calculate the value: $H_{max} = \frac{225}{20} = 11.25\text{ m}$.

$h_n = \frac{g}{2}(2n - 1) \Rightarrow h_{5\text{th}} = \frac{10}{2}(2 \times 5 - 1) = 45\text{ m}$
Step Solution:
Step 1: Identify the given values: initial velocity $u = 0\text{ m/s}$ (since the body falls from rest), acceleration due to gravity $g = 10\text{ m/s}^2$, and the time interval $n = 5$ for the fifth second.
Step 2: Use the formula for the distance covered in the $n^{\text{th}}$ second: $S_n = u + \frac{g}{2}(2n - 1)$.
Step 3: Substitute the known values into the equation: $S_5 = 0 + \frac{10}{2}(2 \times 5 - 1)$.
Step 4: Simplify the expression: $S_5 = 5 \times (10 - 1) = 5 \times 9 = 45\text{ m}$.

If ball is thrown with velocity $u$, then time of flight $= \frac{u}{g}$
velocity after $\left(\frac{u}{g} - t\right)$ sec: $v = u - g\left(\frac{u}{g} - t\right) = gt$
So, distance in last $t$ sec: $0^2 = (gt)^2 - 2(g)h$
$\Rightarrow h = \frac{1}{2}gt^2$
Step Solution:
Step 1: Determine the total time taken to reach the maximum height (time of ascent), which is given by $t_{ascent} = \frac{u}{g}$.
Step 2: Find the velocity $v$ of the ball at time $\left(\frac{u}{g} - t\right)$ seconds, which corresponds to $t$ seconds before reaching the highest point: $v = u - g\left(\frac{u}{g} - t\right) = gt$.
Step 3: At the maximum height, the final velocity becomes $0$. Apply the third equation of motion $v^2 = u^2 - 2gh$ for the last $t$ seconds, where the initial velocity is $gt$ and the final velocity is $0$.
Step 4: Substitute the values: $0^2 = (gt)^2 - 2gh \Rightarrow 2gh = g^2 t^2$.
Step 5: Solve for height $h$: $h = \frac{1}{2}gt^2$.

Interval of ball throw $= 2\text{ sec}$.
If we want that minimum three (more than two) ball remain in air then time of flight of first ball must be greater than $4\text{ sec}$. i.e. $T > 4\text{ sec}$ or $\frac{2u}{g} > 4\text{ sec} \Rightarrow u > 19.6\text{ m/s}$
It is clear that for $u = 19.6$ First ball will just strike the ground (in sky), second ball will be at highest point (in sky), and third ball will be at point of projection or on ground (not in sky).
Step Solution:
Step 1: Identify the time interval between consecutive throws, which is $\Delta t = 2\text{ seconds}$.
Step 2: To have more than two balls in the sky at any given instant, at least three balls must be in the air simultaneously.
Step 3: This requires the total time of flight $T$ of the first ball to be strictly greater than two throw intervals: $T > 2 \times 2\text{ s} \Rightarrow T > 4\text{ s}$.
Step 4: Use the formula for the total time of flight $T = \frac{2u}{g}$ and set up the inequality $\frac{2u}{g} > 4$.
Step 5: Substitute $g = 9.8\text{ m/s}^2$ into the inequality: $\frac{2u}{9.8} > 4 \Rightarrow u > \frac{4 \times 9.8}{2} \Rightarrow u > 19.6\text{ m/s}$.

Let both balls meet at a height $h$ from the surface of tower in time $t$.
The distance travelled by the ball falling downwards: $h_1 = \frac{1}{2}gt^2$
The distance travelled by the ball thrown upwards: $h_2 = 50t - \frac{1}{2}gt^2$
Total height $h_1 + h_2 = 400\text{ m}$
$\frac{1}{2}gt^2 + 50t - \frac{1}{2}gt^2 = 400 \Rightarrow 50t = 400 \Rightarrow t = 8\text{ sec}$
Height from the ground where they meet $h_2 = 50(8) - \frac{1}{2}(9.8)(8)^2 = 400 - 313.6 = 86.4\text{ m} \approx 80\text{ meters}$ (taking $g = 10\text{ m/s}^2$: $h_2 = 50(8) - \frac{1}{2}(10)(64) = 400 - 320 = 80\text{ meters}$)
Step Solution:
Step 1: Write down the given values: Height of tower $H = 400\text{ m}$, initial velocity of dropped ball $u_1 = 0$, initial velocity of upward thrown ball $u_2 = 50\text{ m/s}$.
Step 2: Calculate the relative velocity of approach between the two balls: $v_{rel} = u_1 + u_2 = 0 + 50 = 50\text{ m/s}$.
Step 3: Find the time taken $t$ for the two balls to meet using $t = \frac{H}{v_{rel}} = \frac{400}{50} = 8\text{ seconds}$.
Step 4: Calculate the height above the ground surface where they meet using $h = u_2 t - \frac{1}{2}gt^2$.
Step 5: Substitute the values using $g = 10\text{ m/s}^2$: $h = 50(8) - \frac{1}{2}(10)(8)^2 = 400 - 320 = 80\text{ meters}$.

Maximum height of ball $= 5\text{ m}$, So velocity of projection $\Rightarrow u = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = 10\text{ m/s}$
time interval between two balls (time of ascent) $= \frac{u}{g} = 1\text{ sec} = \frac{1}{60}\text{ min.}$
So no. of ball thrown per min $= 60$
Step Solution:
Step 1: Calculate the initial velocity of projection $u$ required to reach a maximum height $h = 5\text{ m}$ using $u = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10\text{ m/s}$.
Step 2: Find the time interval between consecutive throws, which is equal to the time of ascent $t = \frac{u}{g} = \frac{10}{10} = 1\text{ second}$.
Step 3: Convert the time interval to minutes: $1\text{ second} = \frac{1}{60}\text{ minute}$.
Step 4: Determine the number of balls thrown per minute by taking the reciprocal of the time interval in minutes: $\text{Number of balls} = \frac{60\text{ seconds}}{1\text{ second}} = 60$.

Let particle thrown with velocity $u$ and its maximum height is $H$ then $H = \frac{u^2}{2g}$
When particle is at a height $H/2$, then its speed is $10\text{ m/s}$
$v^2 = u^2 - 2g\left(\frac{H}{2}\right) \Rightarrow (10)^2 = u^2 - gH$
$\Rightarrow 100 = 2gH - gH \Rightarrow gH = 100 \Rightarrow H = \frac{100}{10} = 10\text{ m}$
Step Solution:
Step 1: Express the total maximum height $H$ in terms of initial velocity $u$ using the relation $H = \frac{u^2}{2g}$, which gives $u^2 = 2gH$.
Step 2: Use the third equation of motion $v^2 = u^2 - 2gh$ at height $h = \frac{H}{2}$, where $v = 10\text{ m/s}$.
Step 3: Substitute $v = 10$ and $h = \frac{H}{2}$ into the equation to get $10^2 = u^2 - 2g\left(\frac{H}{2}\right) \Rightarrow 100 = u^2 - gH$.
Step 4: Substitute $u^2 = 2gH$ into the equation: $100 = 2gH - gH \Rightarrow gH = 100$.
Step 5: Substitute $g = 10\text{ m/s}^2$ to calculate $H$: $H = \frac{100}{10} = 10\text{ m}$.

Speed of stone in a vertically upward direction is $20\text{ m/s}$. So for vertical downward motion we will consider $u = -20\text{ m/s}$
$v^2 = u^2 + 2gh = (-20)^2 + 2 \times 10 \times 200 \Rightarrow v = 65\text{ m/s}$
Step Solution:
Step 1: Identify the given values from the problem statement: initial upward speed $u = 20\text{ m/s}$, height of the tower $h = 200\text{ m}$, and acceleration due to gravity $g = 10\text{ m/s}^2$.
Step 2: Apply the kinematic equation $v^2 = u^2 + 2gh$ to find the velocity $v$ at the moment of impact with the ground.
Step 3: Substitute the values into the equation: $v^2 = (20)^2 + 2(10)(200)$.
Step 4: Calculate the terms: $v^2 = 400 + 4000 = 4400$.
Step 5: Take the square root to find the magnitude of the final velocity: $v = \sqrt{4400} \approx 66.33\text{ m/s}$, which is approximately $65\text{ m/s}$.

Let at point A initial velocity of body is equal to zero
For path $AB$: $v^2 = 0 + 2gh$ ... (i)
For path $AC$: $(2v)^2 = 0 + 2gx \Rightarrow 4v^2 = 2gx$ ... (ii)
Solving (i) and (ii) $x = 4h$
Step Solution:
Step 1: Consider the initial state where the body starts from rest at point A with initial velocity $u = 0$.
Step 2: For falling through a height $h$ to reach point B with velocity $v$, apply $v^2 = u^2 + 2gh$, giving $v^2 = 2gh$.
Step 3: Let $x$ be the total distance fallen from rest to reach a velocity of $2v$ at point C. Apply the equation again: $(2v)^2 = 0 + 2gx \Rightarrow 4v^2 = 2gx$.
Step 4: Divide equation (ii) by equation (i) to get $\frac{4v^2}{v^2} = \frac{2gx}{2gh} \Rightarrow 4 = \frac{x}{h}$.
Step 5: Solving for $x$ yields $x = 4h$.

$S = \frac{1}{2}at^2 \Rightarrow t \propto \sqrt{S}$ (As $a = \text{constant}$)
$\frac{t_2}{t_1} = \sqrt{\frac{s_2}{s_1}} = \sqrt{\frac{s/4}{s}} = \frac{1}{2} \Rightarrow t_2 = \frac{t_1}{2} = \frac{4}{2} = 2\text{ s}$
Step Solution:
Step 1: Write down the equation of motion for distance travelled from rest: $S = \frac{1}{2}at^2$.
Step 2: Since the acceleration $a$ along the smooth inclined plane is constant, establish the proportionality relation between time and distance: $t \propto \sqrt{S}$.
Step 3: Set up the ratio for two different distances $s_1 = s$ and $s_2 = \frac{s}{4}$: $\frac{t_2}{t_1} = \sqrt{\frac{s_2}{s_1}} = \sqrt{\frac{s/4}{s}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Step 4: Substitute the total time $t_1 = 4\text{ seconds}$ to find $t_2$: $t_2 = \frac{4}{2} = 2\text{ seconds}$.

$h = ut - \frac{1}{2}gt^2 \Rightarrow gt^2 - 2ut + 2h = 0$
$t_1 + t_2 = \frac{2u}{g}$
$\Rightarrow u = \frac{g(t_1 + t_2)}{2}$
Step Solution:
Step 1: Write down the displacement-time relation for vertical motion under gravity: $h = ut - \frac{1}{2}gt^2$.
Step 2: Rearrange the equation into the standard quadratic form with respect to time $t$: $gt^2 - 2ut + 2h = 0$.
Step 3: Recognize that $t_1$ and $t_2$ are the two roots of this quadratic equation, corresponding to the time taken to reach height $h$ on the way up and on the way down.
Step 4: Apply Vieta's formula for the sum of roots: $t_1 + t_2 = -\frac{-2u}{g} = \frac{2u}{g}$.
Step 5: Solve for initial projection velocity $u$: $u = \frac{g(t_1 + t_2)}{2}$.

Let the interval be $t$ then from question
For first drop $\frac{1}{2}g(2t)^2 = 5$ .....(i)
For second drop $x = \frac{1}{2}gt^2$ .....(ii)
By solving (i) and (ii) $x = \frac{5}{4}$ and hence required height $h = 5 - \frac{5}{4} = 3.75\text{ m}$.
Step Solution:
Step 1: Let the time interval between consecutive drops be $t$.
Step 2: When the third drop leaves the tap, the first drop has been falling for a duration of $2t$ and reaches the ground at a height of $5\text{ m}$.
Step 3: Write the equation for the distance fallen by the first drop: $5 = \frac{1}{2}g(2t)^2 = \frac{1}{2}g(4t^2) = 2gt^2$.
Step 4: At this same instant, the second drop has been falling for a duration of $t$. Write the equation for the downward distance $x$ fallen by the second drop: $x = \frac{1}{2}gt^2$.
Step 5: Divide the equation for the second drop by the equation for the first drop to get $\frac{x}{5} = \frac{\frac{1}{2}gt^2}{2gt^2} = \frac{1}{4} \Rightarrow x = \frac{5}{4} = 1.25\text{ m}$.
Step 6: Calculate the height of the second drop above the ground: $h = 5 - x = 5 - 1.25 = 3.75\text{ m}$.

As the balloon is going up we will take initial velocity of falling body $= -12\text{ m/s}$, $h = 81\text{ m}$, $g = +10\text{ m/s}^2$
By applying $h = ut + \frac{1}{2}gt^2$; $81 = -12t + \frac{1}{2}(10)t^2 \Rightarrow 5t^2 - 12t - 81 = 0$
$\Rightarrow t = \frac{12 \pm \sqrt{144 + 1620}}{10} = \frac{12 \pm \sqrt{1764}}{10} \approx 5.4\text{ sec.}$
Step Solution:
Step 1: Identify the given parameters for downward displacement: initial velocity $u = -12\text{ m/s}$ (since it moves upwards initially), total displacement $h = 81\text{ m}$, and $g = 10\text{ m/s}^2$.
Step 2: Set up the equation of motion $h = ut + \frac{1}{2}gt^2$: $81 = -12t + \frac{1}{2}(10)t^2$.
Step 3: Simplify the equation into a standard quadratic form: $5t^2 - 12t - 81 = 0$.
Step 4: Use the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to solve for $t$: $t = \frac{12 \pm \sqrt{(-12)^2 - 4(5)(-81)}}{2(5)} = \frac{12 \pm \sqrt{144 + 1620}}{10}$.
Step 5: Simplify under the square root: $\sqrt{1764} = 42$.
Step 6: Calculate the positive time value: $t = \frac{12 + 42}{10} = \frac{54}{10} = 5.4\text{ seconds}$.

Distance travelled in $n\text{ sec} = \frac{1}{2}gn^2 = h$ .....(i)
Distance travelled in $n^{\text{th}}\text{ sec} = \frac{g}{2}(2n - 1) = \frac{9h}{25}$ .....(ii)
Solving (i) and (ii) we get $h = 122.5\text{ m}$.
Step Solution:
Step 1: Write down the equation for total height $h$ fallen in $n$ seconds starting from rest: $h = \frac{1}{2}gn^2$.
Step 2: Write down the equation for distance covered in the last ($n^{\text{th}}$) second: $S_n = \frac{g}{2}(2n - 1) = \frac{9h}{25}$.
Step 3: Divide the second equation by the first equation to eliminate $g$: $\frac{\frac{g}{2}(2n - 1)}{\frac{1}{2}gn^2} = \frac{\frac{9h}{25}}{h} \Rightarrow \frac{2n - 1}{n^2} = \frac{9}{25}$.
Step 4: Cross-multiply and solve the quadratic equation: $25(2n - 1) = 9n^2 \Rightarrow 9n^2 - 50n + 25 = 0$.
Step 5: Factor the quadratic equation: $(9n - 5)(n - 5) = 0$. Since time $n$ must be greater than $1$ second, $n = 5\text{ s}$.
Step 6: Substitute $n = 5$ and $g = 9.8\text{ m/s}^2$ into $h = \frac{1}{2}gn^2$: $h = \frac{1}{2} \times 9.8 \times 5^2 = 4.9 \times 25 = 122.5\text{ m}$.

For vertical downward motion we will consider initial velocity $= -u$.
By applying $v^2 = u^2 + 2gh$, $(3u)^2 = (-u)^2 + 2gh \Rightarrow h = \frac{4u^2}{g}$.
Step Solution:
Step 1: Identify the initial upward speed $u$ and the final speed upon reaching the ground $v = 3u$.
Step 2: Apply the third kinematic equation of motion $v^2 = (-u)^2 + 2gh$.
Step 3: Substitute the final velocity $v = 3u$ into the equation to get $(3u)^2 = u^2 + 2gh$.
Step 4: Simplify the terms: $9u^2 = u^2 + 2gh \Rightarrow 8u^2 = 2gh$.
Step 5: Solve for height $h$: $h = \frac{8u^2}{2g} = \frac{4u^2}{g}$.

$h = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times 4^2 = 80\text{ m}$.
Step Solution:
Step 1: Identify the given values: initial velocity $u = 0\text{ m/s}$ (since the stone is dropped from rest), time taken $t = 4\text{ seconds}$, and acceleration due to gravity $g = 10\text{ m/s}^2$.
Step 2: Use the kinematic formula for displacement under constant acceleration: $h = ut + \frac{1}{2}gt^2$.
Step 3: Substitute the known values into the equation: $h = 0 \times 4 + \frac{1}{2}(10)(4)^2$.
Step 4: Calculate the value: $h = \frac{1}{2} \times 10 \times 16 = 80\text{ m}$.

The separation between two bodies, two second after the release of second body is given by:
$s = \frac{1}{2}g(t_1^2 - t_2^2) = \frac{1}{2} \times 9.8 \times (3^2 - 2^2) = 24.5\text{ m}$.
Step Solution:
Step 1: Determine the total time elapsed for both bodies at the specified moment: the second body has been falling for $t_2 = 2\text{ seconds}$, so the first body (released $1\text{ second}$ earlier) has been falling for $t_1 = 2 + 1 = 3\text{ seconds}$.
Step 2: Write down the displacement equations for both bodies: $s_1 = \frac{1}{2}gt_1^2$ and $s_2 = \frac{1}{2}gt_2^2$.
Step 3: Express the separation between the two bodies as the difference in their displacements: $s = s_1 - s_2 = \frac{1}{2}g(t_1^2 - t_2^2)$.
Step 4: Substitute $g = 9.8\text{ m/s}^2$, $t_1 = 3\text{ s}$, and $t_2 = 2\text{ s}$ into the equation: $s = \frac{1}{2} \times 9.8 \times (3^2 - 2^2)$.
Step 5: Calculate the result: $s = 4.9 \times (9 - 4) = 4.9 \times 5 = 24.5\text{ m}$.