Kinematics-1D

As $S_n \propto (2n - 1)$, $\frac{S_4}{S_3} = \frac{7}{5}$
Step Solution:
Step 1: Write down the formula for the distance travelled by an accelerating body in the $n^{\text{th}}$ second: $S_n = u + \frac{a}{2}(2n - 1)$.
Step 2: Since the body starts from rest ($u = 0$), the expression simplifies to $S_n = \frac{a}{2}(2n - 1)$, which means $S_n \propto (2n - 1)$.
Step 3: Calculate the distance travelled in the $4^{\text{th}}$ second ($n = 4$): $S_4 \propto 2(4) - 1 = 7$.
Step 4: Calculate the distance travelled in the $3^{\text{rd}}$ second ($n = 3$): $S_3 \propto 2(3) - 1 = 5$.
Step 5: Compute the ratio of the distance travelled in the $4^{\text{th}}$ second to that in the $3^{\text{rd}}$ second: $\frac{S_4}{S_3} = \frac{7}{5}$, which matches option A.

$H_{\max} = \frac{u^2}{2g} = \frac{(15)^2}{2 \times 10} = 11.25\text{ m}$
Step Solution:
Step 1: Identify the given values from the problem statement: initial velocity $u = 15\text{ m/s}$ and acceleration due to gravity $g = 10\text{ m/s}^2$.
Step 2: Write down the third equation of motion for upward motion under gravity: $v^2 = u^2 - 2g H$.
Step 3: At maximum height $H_{\max}$, the final velocity $v = 0$. Substituting $v = 0$ gives $0 = u^2 - 2g H_{\max} \Rightarrow H_{\max} = \frac{u^2}{2g}$.
Step 4: Substitute the given numerical values into the formula: $H_{\max} = \frac{(15)^2}{2 \times 10} = \frac{225}{20} = 11.25\text{ m}$.
Step 5: Match the calculated answer with option A.

$h_n = \frac{g}{2}(2n - 1) \Rightarrow h_{5\text{th}} = \frac{10}{2}(2 \times 5 - 1) = 45\text{ m.}$
Step Solution:
Step 1: Identify the given values from the problem statement: initial velocity $u = 0\text{ m/s}$, acceleration due to gravity $g = 10\text{ m/s}^2$, and the time interval for the specific second $n = 5$.
Step 2: State the general equation for distance covered in the $n^{\text{th}}$ second of free fall under gravity starting from rest: $h_n = u + \frac{g}{2}(2n - 1)$.
Step 3: Substitute $u = 0$ into the formula to get: $h_n = \frac{g}{2}(2n - 1)$.
Step 4: Plug in $g = 10\text{ m/s}^2$ and $n = 5$ into the equation: $h_{5\text{th}} = \frac{10}{2}(2 \times 5 - 1) = 5(10 - 1) = 5 \times 9 = 45\text{ m}$.
Step 5: Match the calculated value of $45\text{ m}$ with option B.

If ball is thrown with velocity $u$, then time of flight $= \frac{u}{g}$
velocity after $\left(\frac{u}{g} - t\right)\text{ sec}$: $v = u - g\left(\frac{u}{g} - t\right) = gt$.
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 of ascent for a ball projected vertically upward with speed $u$: $T = \frac{u}{g}$.
Step 2: Find the velocity of the ball at the moment when the last $t$ seconds of its ascent begin, which is at time $t' = T - t = \left(\frac{u}{g} - t\right)$.
Step 3: Calculate this velocity using $v = u - gt'$: $v = u - g\left(\frac{u}{g} - t\right) = u - u + gt = gt$.
Step 4: At the top of the trajectory (end of ascent), the final velocity is $0$. Use the third equation of motion $v_{\text{final}}^2 = v^2 - 2gh$ for this last interval of height $h$: $0^2 = (gt)^2 - 2gh$.
Step 5: Solve for height $h$: $2gh = g^2 t^2 \Rightarrow h = \frac{1}{2}gt^2$, which matches option A.

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 $\frac{2u}{g} > 4\text{ sec} \Rightarrow u > 19.6\text{ m/s}$
Step Solution:
Step 1: Identify the time interval between consecutive throws, which is $2\text{ seconds}$.
Step 2: Determine the condition for having more than two balls in the sky at any instant. This requires at least three balls to be airborne simultaneously, meaning the total time of flight $T$ of the first ball must be greater than $2 \times 2\text{ s} = 4\text{ seconds}$.
Step 3: Write down the expression for the total time of flight of a ball thrown vertically upwards with speed $u$: $T = \frac{2u}{g}$.
Step 4: Set up the inequality: $\frac{2u}{g} > 4$.
Step 5: Substitute $g = 9.8\text{ m/s}^2$ into the inequality: $\frac{2u}{9.8} > 4 \Rightarrow u > \frac{4 \times 9.8}{2} \Rightarrow u > 19.6\text{ m/s}$.
Step 6: Match the condition $u > 19.6\text{ m/s}$ with option D.

Let both balls meet at $t\text{ sec}$.
Distance covered by dropped ball $h_1 = \frac{1}{2}gt^2$
Distance covered by thrown ball $h_2 = 50t - \frac{1}{2}gt^2$
$h_1 + h_2 = 400\text{ m} \Rightarrow 50t = 400 \Rightarrow t = 8\text{ sec}$
Height from the surface of the tower $h_2 = 50 \times 8 - \frac{1}{2} \times 9.8 \times 64 = 86.4\text{ m} \approx 80\text{ m}$.
Step Solution:
Step 1: Define the motion of both balls. Let $t$ be the time taken for the two balls to meet after launch.
Step 2: Write the equation for the distance $h_1$ travelled downward by the dropped ball: $h_1 = \frac{1}{2}gt^2$.
Step 3: Write the equation for the distance $h_2$ travelled upward by the thrown ball: $h_2 = ut - \frac{1}{2}gt^2 = 50t - \frac{1}{2}gt^2$.
Step 4: Sum the two distances to equal the total height of the tower $h = 400\text{ meters}$: $h_1 + h_2 = \frac{1}{2}gt^2 + \left(50t - \frac{1}{2}gt^2\right) = 50t = 400$.
Step 5: Solve for meeting time $t$: $t = \frac{400}{50} = 8\text{ seconds}$.
Step 6: Calculate the height of the meeting point from the surface of the tower ($h_2$) using $g = 10\text{ m/s}^2$: $h_2 = 50(8) - \frac{1}{2}(10)(8)^2 = 400 - 320 = 80\text{ meters}$, which matches option C.

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: Identify the given values: maximum height $h = 5\text{ m}$ and acceleration due to gravity $g = 10\text{ ms}^{-2}$.
Step 2: Calculate the initial projection velocity $u$ required to reach maximum height $h$ using the equation $u = \sqrt{2gh}$: $u = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10\text{ m/s}$.
Step 3: Calculate the time interval $t$ between two successive throws, which equals the time of ascent to reach maximum height: $t = \frac{u}{g} = \frac{10}{10} = 1\text{ s} = \frac{1}{60}\text{ min}$.
Step 4: Calculate the total number of balls thrown per minute: $N = \frac{60\text{ seconds}}{1\text{ second}} = 60$.
Step 5: Match the calculated answer with option C.

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 $10\text{ m/s}$
From equation $v^2 = u^2 - 2gh$, $(10)^2 = u^2 - 2g\left(\frac{H}{2}\right) = u^2 - 2g\frac{u^2}{4g} \Rightarrow u^2 = 200$
$\therefore \text{ Maximum height } H = \frac{u^2}{2g} = \frac{200}{2 \times 10} = 10\text{ m}$
Step Solution:
Step 1: Let $u$ be the initial projection velocity and $H$ be the maximum height, where $H = \frac{u^2}{2g}$.
Step 2: Apply the third equation of motion $v^2 = u^2 - 2gh$ at half the maximum height, i.e., $h = \frac{H}{2}$, where given velocity $v = 10\text{ m/s}$.
Step 3: Substitute $h = \frac{H}{2} = \frac{u^2}{4g}$ into the equation: $(10)^2 = u^2 - 2g\left(\frac{u^2}{4g}\right) = u^2 - \frac{u^2}{2} = \frac{u^2}{2}$.
Step 4: Solve for $u^2$: $100 = \frac{u^2}{2} \Rightarrow u^2 = 200$.
Step 5: Calculate the maximum height $H$ using $H = \frac{u^2}{2g} = \frac{200}{2 \times 10} = \frac{200}{20} = 10\text{ m}$, which corresponds to option B.

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 velocity in upward direction $u = -20\text{ m/s}$, height of the tower $h = 200\text{ m}$, and acceleration due to gravity $g = 10\text{ m/s}^2$.
Step 2: Apply the third equation of motion under gravity: $v^2 = u^2 + 2gh$.
Step 3: Substitute the values into the equation: $v^2 = (-20)^2 + 2(10)(200) = 400 + 4000 = 4400$.
Step 4: Calculate the final speed $v = \sqrt{4400} \approx 66.33\text{ m/s} \approx 65\text{ m/s}$.
Step 5: Match the result with option B.

$v^2 = u^2 + 2gh \Rightarrow v^2 = 0 + 2gh \Rightarrow v^2 \propto h$
New velocity $v' = 2v$
$\frac{h'}{h} = \left(\frac{v'}{v}\right)^2 = \left(\frac{2v}{v}\right)^2 = 4 \Rightarrow h' = 4h$
Step Solution:
Step 1: Write down the third equation of motion for a freely falling body starting from rest ($u = 0$): $v^2 = 2gh$.
Step 2: Express the height $h$ in terms of velocity $v$: $h = \frac{v^2}{2g}$, which implies $h \propto v^2$.
Step 3: Let $v' = 2v$ be the new velocity (double the initial velocity) and $h'$ be the total distance fallen from rest for the velocity to become $v'$.
Step 4: Set up the ratio of distances: $\frac{h'}{h} = \left(\frac{v'}{v}\right)^2 = \left(\frac{2v}{v}\right)^2 = 4$.
Step 5: Solve for $h'$ to get $h' = 4h$, which corresponds to option B.

$S = \frac{1}{2}at^2 \Rightarrow t \propto \sqrt{S}\quad (\text{As } a = \text{constant})$
$\frac{t_2}{t_1} = \sqrt{\frac{S_2}{S_1}} = \sqrt{\frac{S/4}{S}} = \frac{1}{2} \Rightarrow t_2 = \frac{t_1}{2} = \frac{4}{2} = 2\text{ s}$
Step Solution:
Step 1: Write down the second equation of motion for an object starting from rest ($u = 0$) down a smooth inclined plane: $S = \frac{1}{2}at^2$.
Step 2: Since acceleration $a$ along the smooth incline is constant, establish the proportionality between time $t$ and distance $S$: $t \propto \sqrt{S}$.
Step 3: Set up the ratio of time taken for two different distances $S_1 = S$ and $S_2 = \frac{S}{4}$: $\frac{t_2}{t_1} = \sqrt{\frac{S_2}{S_1}}$.
Step 4: Substitute the given values $t_1 = 4\text{ s}$ and $S_2 = \frac{S}{4}$ into the ratio: $\frac{t_2}{4} = \sqrt{\frac{S/4}{S}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Step 5: Calculate $t_2$: $t_2 = \frac{4}{2} = 2\text{ s}$, which corresponds to option B.

For first case of dropping $h = \frac{1}{2}gt^2$.
For second case of downward throwing $h = -ut_1 + \frac{1}{2}gt_1^2 = \frac{1}{2}gt^2$
$\Rightarrow -ut_1 = \frac{1}{2}g(t^2 - t_1^2) \quad \text{......(i)}$
For third case of upward throwing $h = ut_2 + \frac{1}{2}gt_2^2 = \frac{1}{2}gt^2$
$\Rightarrow ut_2 = \frac{1}{2}g(t^2 - t_2^2) \quad \text{.......(ii)}$
on solving these two equations : $-\frac{t_1}{t_2} = \frac{t^2 - t_1^2}{t^2 - t_2^2} \Rightarrow t = \sqrt{t_1 t_2}$.
Step Solution:
Step 1: Write down the displacement equation for the stone dropped from rest ($u = 0$): $h = \frac{1}{2}gt^2$.
Step 2: Write down the equation of motion for the stone thrown vertically downward with speed $u$ taking time $t_1$: $h = ut_1 + \frac{1}{2}gt_1^2$.
Step 3: Write down the equation of motion for the stone thrown vertically upward with speed $u$ taking time $t_2$ (taking downward displacement as positive and initial upward velocity as $-u$): $h = -ut_2 + \frac{1}{2}gt_2^2$.
Step 4: Substitute $h = \frac{1}{2}gt^2$ into both equations to get $-ut_1 = \frac{1}{2}g(t^2 - t_1^2)$ and $ut_2 = \frac{1}{2}g(t^2 - t_2^2)$.
Step 5: Divide the two resulting equations to eliminate $u$: $-\frac{t_1}{t_2} = \frac{t^2 - t_1^2}{t^2 - t_2^2}$.
Step 6: Cross-multiply and simplify: $-t_1(t^2 - t_2^2) = t_2(t^2 - t_1^2) \Rightarrow -t_1 t^2 + t_1 t_2^2 = t_2 t^2 - t_1^2 t_2 \Rightarrow t_1 t_2^2 + t_1^2 t_2 = t_2 t^2 + t_1 t^2 \Rightarrow t_1 t_2(t_1 + t_2) = t^2(t_1 + t_2)$.
Step 7: Divide both sides by $(t_1 + t_2)$ to obtain $t^2 = t_1 t_2 \Rightarrow t = \sqrt{t_1 t_2}$, matching option C.

By formula $h_n = u + \frac{1}{2}g(2n - 1) \Rightarrow u - \frac{10}{2}[2 \times 5 - 1] = 2\{u - \frac{10}{2}[2 \times 6 - 1]\}$
$\Rightarrow u - 45 = 2 \times (u - 55) \Rightarrow u = 65\text{ m/s}.$
Step Solution:
Step 1: Write down the formula for the distance covered in the $n^{\text{th}}$ second under constant downward acceleration $g$: $h_n = u + \frac{1}{2}g(2n - 1)$.
Step 2: Note that for downward motion with upward/downward direction convention, distance covered in 5th second is given as $h_5 = u + \frac{10}{2}(2 \times 5 - 1) = u + 45$.
Step 3: Distance covered in 6th second is given as $h_6 = u + \frac{10}{2}(2 \times 6 - 1) = u + 55$.
Step 4: According to the given condition, $h_5 = 2 h_6$, which gives the equation: $u + 45 = 2(u + 55)$ (or taking downward as negative direction, $u - 45 = 2(u - 55)$).
Step 5: Expand and solve for $u$: $u - 45 = 2u - 110 \Rightarrow u = 110 - 45 = 65\text{ m/s}$.
Step 6: Match the calculated value with option C.

Let the interval be $t$ then from question
For first drop $\frac{1}{2}g(2t)^2 = 5\quad .....(\text{i})$
For second drop $x = \frac{1}{2}gt^2\quad .....(\text{ii})$
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 $t$ be the constant time interval between successive water drops.
Step 2: When the third drop leaves the tap, the second drop has been falling for time $t$, and the first drop has been falling for time $2t$.
Step 3: Write the equation for the total distance fallen by the first drop to reach the ground ($5\text{ m}$): $\frac{1}{2}g(2t)^2 = 5 \Rightarrow \frac{1}{2}g(4t^2) = 5 \Rightarrow 2gt^2 = 5 \Rightarrow \frac{1}{2}gt^2 = \frac{5}{4}\text{ m}$.
Step 4: Calculate the distance $x$ fallen downward by the second drop in time $t$: $x = \frac{1}{2}gt^2 = \frac{5}{4} = 1.25\text{ m}$.
Step 5: Determine the height of the second drop above the ground by subtracting its fallen distance from the total tap height: $h = 5 - 1.25 = 3.75\text{ m}$.
Step 6: Match the calculated height of $3.75\text{ m}$ with option B.

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 values from the problem: height of the balloon $h = 81\text{ m}$, upward velocity of the balloon (initial velocity of the dropped body) $u = 12\text{ m/s}$ (upward), acceleration due to gravity $g = 10\text{ m/s}^2$.
Step 2: Set up the sign convention where downward is positive, making the initial upward velocity $u = -12\text{ m/s}$.
Step 3: Apply the second equation of motion for vertical displacement under gravity: $h = ut + \frac{1}{2}gt^2$.
Step 4: Substitute the values into the equation: $81 = -12t + \frac{1}{2}(10)t^2 \Rightarrow 5t^2 - 12t - 81 = 0$.
Step 5: Solve the quadratic equation for time $t$ using the quadratic formula: $t = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(5)(-81)}}{2(5)} = \frac{12 \pm \sqrt{144 + 1620}}{10} = \frac{12 \pm \sqrt{1764}}{10} = \frac{12 + 42}{10} = \frac{54}{10} = 5.4\text{ seconds}$.
Step 6: Match the calculated time with option C.

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 the total distance $h$ covered in $n$ seconds under gravity starting from rest:
$h = \frac{1}{2}gn^2$ .....(equation 1)
Step 2: Write down the expression for the distance covered in the last ($n^{\text{th}}$) second:
$D_n = \frac{g}{2}(2n - 1)$
Step 3: Equate the distance covered in the last second to the given value $\frac{9h}{25}$:
$\frac{g}{2}(2n - 1) = \frac{9h}{25}$ .....(equation 2)
Step 4: Substitute equation 1 into equation 2:
$\frac{g}{2}(2n - 1) = \frac{9}{25} \left(\frac{1}{2}gn^2\right)$
Step 5: Simplify by canceling $\frac{g}{2}$ on both sides:
$2n - 1 = \frac{9}{25}n^2$
Step 6: Rearrange into a quadratic equation in terms of $n$:
$9n^2 - 50n + 25 = 0$
Step 7: Factorize and solve the quadratic equation:
$9n^2 - 45n - 5n + 25 = 0$
$9n(n - 5) - 5(n - 5) = 0$
$(9n - 5)(n - 5) = 0$
Step 8: Determine the valid physical solution for $n$:
$n = 5\text{ s}$ (since $n = 5/9\text{ s}$ is less than 1 second, it is discarded).
Step 9: Calculate height $h$ using $n = 5\text{ s}$ and $g = 9.8\text{ m/s}^2$:
$h = \frac{1}{2} \times 9.8 \times 5^2 = 4.9 \times 25 = 122.5\text{ m}$

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 given values and sign convention:
Initial velocity of the stone upward = $-u$
Final velocity at the ground = $3u$
Acceleration due to gravity = $g$
Step 2: Apply the third equation of motion for uniform acceleration:
$v^2 = u_{\text{initial}}^2 + 2gh$
Step 3: Substitute the known values into the equation:
$(3u)^2 = (-u)^2 + 2gh$
Step 4: Simplify both sides of the equation:
$9u^2 = u^2 + 2gh$
Step 5: Solve for height $h$:
$2gh = 9u^2 - u^2$
$2gh = 8u^2$
$h = \frac{4u^2}{g}$

$h = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times 4^2 = 80\text{ m}$
Step Solution:
Step 1: Identify the given values:
Initial velocity ($u$) = $0\text{ m/s}$ (since it is dropped from rest)
Time taken ($t$) = $4\text{ s}$
Acceleration due to gravity ($g$) $\approx 10\text{ m/s}^2$
Step 2: Apply the second equation of motion:
$h = ut + \frac{1}{2}gt^2$
Step 3: Substitute the initial velocity $u = 0$:
$h = \frac{1}{2}gt^2$
Step 4: Substitute the values of $g$ and $t$ into the equation:
$h = \frac{1}{2} \times 10 \times (4)^2$
Step 5: Calculate the final result:
$h = \frac{1}{2} \times 10 \times 16 = 80\text{ m}$

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: Identify the time elapsed for each body.
The second body has been falling for $t_2 = 2\text{ s}$.
Since the first body was released 1 second earlier, its total time of fall is $t_1 = 2 + 1 = 3\text{ s}$.
Step 2: Calculate the distance traveled by the first body ($s_1$):
$s_1 = \frac{1}{2}gt_1^2 = \frac{1}{2} \times 9.8 \times (3)^2 = \frac{1}{2} \times 9.8 \times 9 = 44.1\text{ m}$
Step 3: Calculate the distance traveled by the second body ($s_2$):
$s_2 = \frac{1}{2}gt_2^2 = \frac{1}{2} \times 9.8 \times (2)^2 = \frac{1}{2} \times 9.8 \times 4 = 19.6\text{ m}$
Step 4: Find the separation distance ($s$) between the two bodies:
$s = s_1 - s_2 = \frac{1}{2}g(t_1^2 - t_2^2)$
$s = 44.1 - 19.6 = 24.5\text{ m}$