Motion in 1D
38 Questions
Start DPT Test
Q1
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Displacement is a vector quantity defined as the shortest distance between the initial position and the final position. When a body moves in three mutually perpendicular directions, the magnitude of the net displacement vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $r = \sqrt{x^2 + y^2 + z^2}$.
A body moves $6\text{ m}$ north, $8\text{ m}$ east and $10\text{ m}$ vertically upwards. What is its resultant displacement from initial position?
A.
$10\sqrt{2}\text{ m}$
B.
$10\text{ m}$
C.
$\frac{10}{\sqrt{2}}\text{ m}$
D.
$10 \times 2\text{ m}$
Q2
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Displacement is a vector quantity representing the shortest straight-line distance from the initial position to the final position. When two displacement vectors are perpendicular to each other, the magnitude of the resultant displacement vector $\vec{r} = x\hat{i} + y\hat{j}$ is calculated using the Pythagorean theorem: $r = \sqrt{x^2 + y^2}$.
A man goes $10\text{ m}$ towards North, then $20\text{ m}$ towards east then displacement is
A.
$30\text{ m}$
B.
$25.5\text{ m}$
C.
$22.5\text{ m}$
D.
$25\text{ m}$
Q3
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Displacement is a vector quantity representing the shortest path from the initial to the final position. A motion in a plane can be broken down into orthogonal vector components along the East-West (x-axis) and North-South (y-axis) directions. The total displacement is the vector sum of individual displacement vectors $\vec{r} = \vec{r}_1 + \vec{r}_2 + \vec{r}_3 = x\hat{i} + y\hat{j}$.
A person moves $30\text{ m}$ north and then $20\text{ m}$ towards east and finally $30\sqrt{2}\text{ m}$ in south-west direction. The displacement of the person from the origin will be
A.
$10\text{ m}$ along north
B.
$10\text{ m}$ along south
C.
$10\text{ m}$ along west
D.
Zero
Q4
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Displacement is a vector quantity defined as the shortest distance from the initial position to the final position. In three-dimensional space, the net displacement vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is obtained by summing the individual vector displacements along mutually perpendicular axes. The magnitude of the displacement vector is given by $r = \sqrt{x^2 + y^2 + z^2}$.
An aeroplane flies $400\text{ m}$ north and $300\text{ m}$ south and then flies $1200\text{ m}$ upwards then net displacement is
A.
$1400\text{ m}$
B.
$1200\text{ m}$
C.
$1300\text{ m}$
D.
$1500\text{ m}$
Q5
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Displacement is a vector quantity defined as the shortest straight-line distance from the initial position to the final position. For motion along a circular path, completing one full revolution brings the object back to its starting point, resulting in zero displacement for that complete turn. If an object stops at a diametrically opposite point after a fractional revolution, the magnitude of the displacement is equal to the diameter of the circular track ($2R$).
An athlete completes one round of a circular track of radius $R$ in $40\text{ sec}$. What will be his displacement at the end of $2\text{ min } 20\text{ sec}$?
A.
$2\pi R$
B.
$7\pi R$
C.
$2R$
D.
Zero
Q6
DPT
DPT-1
MCQ
25 Jul 2026
Concept: When a circular wheel rolls forward without slipping, the point initially in contact with the ground undergoes both horizontal translation and vertical motion. During half a revolution, the horizontal displacement of the point is equal to half the circumference of the wheel ($\pi r$), and its vertical displacement from the bottom to the top position is equal to the diameter of the wheel ($2r$). The net displacement magnitude is calculated using the Pythagorean theorem: $S = \sqrt{x^2 + y^2}$.
A wheel of radius $1\text{ meter}$ rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is
A.
$2\pi$
B.
$\sqrt{2}\pi$
C.
$\sqrt{\pi^2 + 4}$
D.
$\pi$
Q7
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Average velocity is defined as the total displacement divided by the total time taken. When a journey is divided into equal distance intervals covered at different uniform speeds, the average velocity is given by the harmonic mean of the speeds.
A person travels along a straight road for half the distance with velocity $v_1$ and the remaining half distance with velocity $v_2$. The average velocity is given by
A.
$v_1 v_2$
B.
$\frac{v_2^2}{v_1^2}$
C.
$\frac{v_1 + v_2}{2}$
D.
$\frac{2 v_1 v_2}{v_1 + v_2}$
Q8
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Average speed is defined as the total distance covered divided by the total time taken. When a body travels equal distances at two different speeds $v_1$ and $v_2$, the average speed is given by the harmonic mean of the two speeds: $v_{av} = \frac{2 v_1 v_2}{v_1 + v_2}$.
A car travels from $A$ to $B$ at a speed of $20\text{ km/hr}$ and returns at a speed of $30\text{ km/hr}$. The average speed of the car for the whole journey is
A.
$25\text{ km/hr}$
B.
$24\text{ km/hr}$
C.
$50\text{ km/hr}$
D.
$5\text{ km/hr}$
Q9
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Average speed is defined as the total distance covered divided by the total time taken for the journey. When a body covers equal distances for the forward and return journeys at different uniform speeds $v_1$ and $v_2$, the average speed is given by $v_{av} = \frac{2 v_1 v_2}{v_1 + v_2}$.
A boy walks to his school at a distance of $6\text{ km}$ with constant speed of $2.5\text{ km/hr}$ and walks back with a constant speed of $4\text{ km/hr}$. His average speed for round trip expressed in $\text{km/hour}$ is
A.
$3$
B.
$1/2$
C.
$24/13$
D.
$40/13$
Q10
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Average speed is defined as the total distance covered divided by the total time taken for the journey. When a journey is divided into two equal distances covered at different uniform speeds $v_1$ and $v_2$, the average speed is given by the harmonic mean of the speeds: $v_{av} = \frac{2 v_1 v_2}{v_1 + v_2}$.
A car travels the first half of a distance between two places at a speed of $30\text{ km/hr}$ and the second half of the distance at $50\text{ km/hr}$. The average speed of the car for the whole journey is
A.
$40.0\text{ km/hr}$
B.
$42.5\text{ km/hr}$
C.
$37.5\text{ km/hr}$
D.
$35.0\text{ km/hr}$
Q11
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Average speed is defined as the total distance covered divided by the total time taken for the trip. When a journey is divided into unequal distance segments traveled at different constant speeds, the total time is calculated by summing the time taken for each segment $t_i = \frac{s_i}{v_i}$. The average speed is then $v_{av} = \frac{S}{T_{total}}$.
One car moving on a straight road covers one third of the distance with $20\text{ km/hr}$ and the rest with $60\text{ km/hr}$. The average speed is
A.
$40\text{ km/hr}$
B.
$36\text{ km/hr}$
C.
$46\frac{2}{3}\text{ km/hr}$
D.
$80\text{ km/hr}$
Q12
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Average speed is defined as total distance divided by total time elapsed. When a body moves with uniform speed $v_1$ for time $t_1$ and uniform speed $v_2$ for time $t_2$, the average speed is given by the time-weighted average $v_{av} = \frac{v_1 t_1 + v_2 t_2}{t_1 + t_2}$. When the time intervals are equal ($t_1 = t_2 = \frac{t}{2}$), the average speed simplifies to the arithmetic mean of the speeds: $v_{av} = \frac{v_1 + v_2}{2}$.
A car moves for half of its time at $80\text{ km/h}$ and for rest half of time at $40\text{ km/h}$. Total distance covered is $60\text{ km}$. What is the average speed of the car?
A.
$60\text{ km/h}$
B.
$80\text{ km/h}$
C.
$120\text{ km/h}$
D.
$180\text{ km/h}$
Q13
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Average speed is defined as the ratio of total distance traveled to total time taken. When a body moves with different speeds over different time intervals, the individual distances covered are calculated as $d = v \times t$. The average speed is then calculated using $v_{av} = \frac{d_1 + d_2}{t_1 + t_2}$.
A train has a speed of $60\text{ km/h}$ for the first one hour and $40\text{ km/h}$ for the next half hour. Its average speed in $\text{km/h}$ is
A.
$50$
B.
$53.33$
C.
$48$
D.
$70$
Q14
DPT
DPT-1
MCQ
25 Jul 2026
Concept: When a train crosses an object with a non-negligible length, such as a bridge, the total distance covered by the train to completely cross it is equal to the sum of the length of the train and the length of the bridge ($D = L_{train} + L_{bridge}$). The speed is converted from $\text{km/h}$ to $\text{m/s}$ using the conversion factor $1\text{ km/h} = \frac{5}{18}\text{ m/s}$, and the time required is determined using $t = \frac{D}{v}$.
A $150\text{ m}$ long train is moving with a uniform velocity of $45\text{ km/h}$. The time taken by the train to cross a bridge of length $850\text{ meters}$ is
A.
$56\text{ sec}$
B.
$68\text{ sec}$
C.
$80\text{ sec}$
D.
$92\text{ sec}$
Q15
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Displacement is a vector quantity defined as the net change in position from the starting point to the ending point. If an object returns to its initial position, its total displacement is zero regardless of the distance traveled. Average speed is a scalar quantity calculated as total distance divided by total time: $v_{av} = \frac{\text{Total distance}}{\text{Total time}}$.
A particle is constrained to move on a straight line path. It returns to the starting point after $10\text{ sec}$. The total distance covered by the particle during this time is $30\text{ m}$. Which of the following statements about the motion of the particle is false?
A.
Displacement of the particle is zero
B.
Average speed of the particle is $3\text{ m/s}$
C.
Displacement of the particle is $30\text{ m}$
D.
Both (a) and (b)
Q16
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Average speed is defined as the total distance covered divided by the total time taken ($v_{av} = \frac{\text{Total distance}}{\text{Total time}}$). When calculating average speed over a specific time interval, one must determine the position and distance traveled up to that exact time limit by considering the separate legs of the motion.
A man walks on a straight road from his home to a market $2.5\text{ km}$ away with a speed of $5\text{ km/h}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5\text{ km/h}$. The average speed of the man over the interval of time $0$ to $40\text{ min}$ is equal to
A.
$\frac{30}{4}\text{ km/h}$
B.
$5\text{ km/h}$
C.
$\frac{25}{4}\text{ km/h}$
D.
$\frac{45}{8}\text{ km/h}$
Q17
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Average velocity is defined as total displacement divided by total time ($\vec{v}_{av} = \frac{\Delta\vec{r}}{\Delta t}$), while average speed is defined as total distance divided by total time ($v_{av} = \frac{\Delta s}{\Delta t}$). Since distance is always greater than or equal to the magnitude of displacement ($\text{Distance} \ge \vert{}\text{Displacement}\vert{}$), the magnitude of average velocity is always less than or equal to the average speed. Therefore, the ratio of average velocity to average speed is unity or less ($\le 1$).
The ratio of the numerical values of the average velocity and average speed of a body is always
A.
Unity
B.
Unity or less
C.
Unity or more
D.
Less than unity
Q18
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Mean (or average) velocity is defined as total displacement divided by total time elapsed. When a body moves along a straight path with different uniform velocities for equal time intervals ($t_1 = t_2 = \frac{T}{2}$), the mean velocity is given by the arithmetic mean of the individual velocities: $V = \frac{v_1 + v_2}{2}$.
A person travels along a straight road for the first half time with a velocity $v_1$ and the next half time with a velocity $v_2$. The mean velocity $V$ of the man is
A.
$\frac{2}{V} = \frac{1}{v_1} + \frac{1}{v_2}$
B.
$V = \frac{v_1 + v_2}{2}$
C.
$V = \sqrt{v_1 v_2}$
D.
$V = \sqrt{\frac{v_1}{v_2}}$
Q19
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Average speed is defined as the total distance covered divided by the total time taken for the trip. When a journey is divided into distance fractions $f_1$ and $f_2$ covered at speeds $v_1$ and $v_2$ respectively, the total time is calculated by adding the time taken for each fraction: $T = \frac{f_1 S}{v_1} + \frac{f_2 S}{v_2}$. Average speed is then given by $v_{av} = \frac{S}{T}$.
If a car covers $2/5$ of the total distance with $v_1$ speed and $3/5$ distance with $v_2$ then average speed is
A.
$\frac{1}{2}\sqrt{v_1 v_2}$
B.
$\frac{v_1 + v_2}{2}$
C.
$\frac{2 v_1 v_2}{v_1 + v_2}$
D.
$\frac{5 v_1 v_2}{3 v_1 + 2 v_2}$
Q20
DPT
DPT-1
MCQ
25 Jul 2026
Concept: Average velocity is defined as the total displacement divided by the total time taken. When a body moves with velocities $v_1, v_2, \dots, v_n$ for equal intervals of time ($t_1 = t_2 = \dots = t_n = t$), the average velocity simplifies to the arithmetic mean of the individual velocities: $v_{av} = \frac{v_1 + v_2 + \dots + v_n}{n}$.
A particle moves for $20\text{ seconds}$ with velocity $3\text{ m/s}$ and then velocity $4\text{ m/s}$ for another $20\text{ seconds}$ and finally moves with velocity $5\text{ m/s}$ for next $20\text{ seconds}$. What is the average velocity of the particle?
A.
$3\text{ m/s}$
B.
$4\text{ m/s}$
C.
$5\text{ m/s}$
D.
$\text{Zero}$
Q21
DPT
DPT-1
MCQ
25 Jul 2026
Concept: When an object covers two equal distances with different uniform velocities $v_1$ and $v_2$, the average velocity over the total distance is given by the harmonic mean of the two velocities:
$v_{avg} = \frac{2 v_1 v_2}{v_1 + v_2}$
A car travels half the distance with constant velocity of 40 km/h and the remaining half with a constant velocity of 60 km/h. The average velocity of the car in km/h is:
$v_{avg} = \frac{2 v_1 v_2}{v_1 + v_2}$
A.
40
B.
45
C.
48
D.
50
Q22
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: When a body is projected vertically upward under gravity, it experiences a constant deceleration equal to $g$. At the maximum height, its final velocity becomes $0$. Using the kinematic equation $v^2 = u^2 - 2g H_{max}$, the formula for maximum height attained is $H_{max} = \frac{u^2}{2g}$.
If a body is thrown up with the velocity of $15\text{ m/s}$ then maximum height attained by the body is ($g = 10\text{ m/s}^2$)
A.
$11.25\text{ m}$
B.
$16.2\text{ m}$
C.
$24.5\text{ m}$
D.
$7.62\text{ m}$
Q23
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: When an object is dropped from rest under gravity, its initial velocity $u = 0$, and it undergoes constant downward acceleration equal to $g$. The distance travelled by a body in the $n^{\text{th}}$ second of uniform acceleration is given by the formula $S_n = u + \frac{g}{2}(2n - 1)$.
A body falls from rest in the gravitational field of the earth. The distance travelled in the fifth second of its motion is ($g = 10\text{ m/s}^2$)
A.
$25\text{ m}$
B.
$45\text{ m}$
C.
$90\text{ m}$
D.
$125\text{ m}$
Q24
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: During vertical upward motion under constant gravitational retardation $g$, the distance covered in the last $t$ seconds of ascent is equivalent to the distance a freely falling body covers in $t$ seconds when starting from rest at the maximum height point. The kinematic formula used is $h = ut + \frac{1}{2}gt^2$ with $u = 0$.
If a ball is thrown vertically upwards with speed $u$, from the image gven below, the distance covered during the last $t$ seconds of its ascent is
A.
$\frac{1}{2}gt^2$
B.
$ut - \frac{1}{2}gt^2$
C.
$(u - gt)t$
D.
$ut$
Q25
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: For more than two balls to remain in the air (sky) simultaneously, the total time of flight $T$ of a single ball must be greater than the combined time interval for throwing the subsequent two balls. The total time of flight for a ball projected vertically upward with speed $u$ under gravity is given by $T = \frac{2u}{g}$.
A man throws balls with the same speed vertically upwards one after the other at an interval of $2\text{ seconds}$. What should be the speed of the throw so that more than two balls are in the sky at any time (Given $g = 9.8\text{ m/s}^2$)
A.
At least $0.8\text{ m/s}$
B.
Any speed less than $19.6\text{ m/s}$
C.
Only with speed $19.6\text{ m/s}$
D.
More than $19.6\text{ m/s}$
Q26
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: When two bodies move towards each other under gravity, the relative acceleration between them is zero ($g - g = 0$). The relative velocity remains constant, equal to the relative initial velocity $v_{rel} = u_1 + u_2$. The time to meet is $t = \frac{h}{v_{rel}}$. The height from the ground where they meet can then be determined using kinematic equations for the upward-moving body: $h = ut - \frac{1}{2}gt^2$.
A man drops a ball downside from the roof of a tower of height $400\text{ meters}$. At the same time another ball is thrown upside with a velocity $50\text{ meter/sec}$ from the surface of the tower, then they will meet at which height from the surface of the tower
A.
$100\text{ meters}$
B.
$320\text{ meters}$
C.
$80\text{ meters}$
D.
$240\text{ meters}$
Q27
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: When a ball is projected vertically upward to reach a maximum height $h$, its initial velocity is given by $u = \sqrt{2gh}$. The time taken to reach the maximum height (time of ascent) is $t = \frac{u}{g} = \sqrt{\frac{2h}{g}}$. The rate of balls thrown per minute is obtained by dividing $60\text{ seconds}$ by the time interval $t$ between consecutive throws.
A very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is $5\text{ m}$, the number of ball thrown per minute is (take $g = 10\text{ ms}^{-2}$)
A.
$120$
B.
$80$
C.
$60$
D.
$40$
Q28
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: When a body is projected vertically upwards under constant gravitational retardation $g$, its velocity $v$ at any height $h$ is given by the kinematic relation $v^2 = u^2 - 2gh$. The maximum height $H$ is attained when the velocity becomes zero ($0 = u^2 - 2gH \Rightarrow H = \frac{u^2}{2g}$). At half the maximum height ($h = \frac{H}{2}$), the equation becomes $v^2 = u^2 - 2g\left(\frac{H}{2}\right) = u^2 - gH$.
A particle is thrown vertically upwards. If its velocity at half of the maximum height is $10\text{ m/s}$, then maximum height attained by it is (Take $g = 10\text{ m/s}^2$)
A.
$8\text{ m}$
B.
$10\text{ m}$
C.
$12\text{ m}$
D.
$16\text{ m}$
Q29
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: When a body is projected upwards from a certain height, its overall displacement when reaching the ground is equal to the negative of the tower's height ($h = -200\text{ m}$). Using the third kinematic equation of motion under gravity, $v^2 = u^2 + 2gh$, we can determine the final velocity with which the body strikes the ground.
A stone is shot straight upward with a speed of $20\text{ m/sec}$ from a tower $200\text{ m}$ high. The speed with which it strikes the ground is approximately
A.
$60\text{ m/sec}$
B.
$65\text{ m/sec}$
C.
$70\text{ m/sec}$
D.
$75\text{ m/sec}$
Q30
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: For a body falling freely from rest under gravity ($u = 0$), the velocity acquired after falling through a distance $x$ is given by the third equation of motion: $v^2 = u^2 + 2gx \Rightarrow v^2 = 2gx$. Therefore, distance fallen is directly proportional to the square of velocity ($x \propto v^2$).
A body freely falling from the rest has a velocity $v$ after it falls through a height $h$. The distance it has to fall down for its velocity to become double, is from the image gven below
A.
$2h$
B.
$4h$
C.
$6h$
D.
$8h$
Q31
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: For a body starting from rest ($u = 0$) and moving under uniform acceleration $a$ along an inclined plane, the distance covered $S$ in time $t$ is given by $S = \frac{1}{2}at^2$. Since $a$ is constant, the time taken is proportional to the square root of the distance covered ($t \propto \sqrt{S}$).
A body sliding on a smooth inclined plane requires $4\text{ seconds}$ to reach the bottom starting from rest at the top. How much time does it take to cover one-fourth distance starting from rest at the top
A.
$1\text{ s}$
B.
$2\text{ s}$
C.
$4\text{ s}$
D.
$16\text{ s}$
Q32
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: For a body projected vertically upward with initial velocity $u$, its height $h$ at time $t$ is given by $h = ut - \frac{1}{2}gt^2$, which rearranges into a quadratic equation in $t$: $gt^2 - 2ut + 2h = 0$. The roots $t_1$ and $t_2$ represent the times at which the body passes height $h$ during upward and downward motion. By Vieta's formulas, the sum of roots is $t_1 + t_2 = \frac{2u}{g}$, which gives $u = \frac{g(t_1 + t_2)}{2}$.
A body is projected vertically upwards with a velocity $u$. It crosses a point at a height $h$ after $t_1$ and $t_2$ seconds respectively. The speed of projection $u$ is
A.
$\frac{g(t_1 + t_2)}{2}$
B.
$g(t_1 + t_2)$
C.
$\frac{g(t_1 - t_2)}{2}$
D.
$g \sqrt{t_1 t_2}$
Q33
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: For a body falling freely under gravity from rest ($u = 0$), the distance fallen in time $t$ is given by $s = \frac{1}{2}gt^2$. When events occur at equal time intervals $t$, the total time elapsed for the first drop when the third drop leaves is $2t$. The distance fallen by the second drop during time $t$ can be found using the ratio of distances covered in proportional times. The height of the drop above the ground is $h = H - s$.
Water drops fall at regular intervals from a tap which is $5\text{ m}$ above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant
A.
$2.50\text{ m}$
B.
$3.75\text{ m}$
C.
$4.00\text{ m}$
D.
$1.25\text{ m}$
Q34
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: When an object is released from an ascending balloon, it inherits the upward velocity of the balloon as its initial velocity ($u = -12\text{ m/s}$ taking downward direction as positive). The motion under gravity is governed by the kinematic equation $h = ut + \frac{1}{2}gt^2$, where $h$ is the total downward displacement to reach the ground.
A balloon is at a height of $81\text{ m}$ and is ascending upwards with a velocity of $12\text{ m/s}$. A body of $2\text{ kg}$ weight is dropped from it. If $g = 10\text{ m/s}^2$, the body will reach the surface of the earth in
A.
$1.5\text{ s}$
B.
$4.025\text{ s}$
C.
$5.4\text{ s}$
D.
$6.75\text{ s}$
Q35
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: For a body dropped from rest ($u = 0$) under gravity, total height $h$ fallen in time $n$ seconds is given by $h = \frac{1}{2}gn^2$. The distance travelled in the $n^{\text{th}}$ second (last second) is given by $S_n = \frac{g}{2}(2n - 1)$. By setting $S_n = \frac{9h}{25}$, we can solve for total time $n$ and subsequently determine the height $h$.
A particle is dropped under gravity from rest from a height $h$ ($g = 9.8\text{ m/s}^2$) and it travels a distance $9h/25$ in the last second, the height $h$ is
A.
$100\text{ m}$
B.
$122.5\text{ m}$
C.
$145\text{ m}$
D.
$167.5\text{ m}$
Q36
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: When an object is thrown vertically upward with speed $u$ from a height $h$, its initial velocity for vertical downward motion can be considered as $-u$ relative to the point of projection. Using the third equation of motion under gravity, $v^2 = u^2 + 2gh$, where $v$ is the speed upon striking the ground, the height of the tower $h$ can be determined.
A stone thrown upward with a speed $u$ from the top of the tower reaches the ground with a velocity $3u$. The height of the tower is
A.
$3u^2 / g$
B.
$4u^2 / g$
C.
$6u^2 / g$
D.
$9u^2 / g$
Q37
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: When a body is dropped from rest under gravity, its initial velocity $u = 0$ and it undergoes constant downward acceleration equal to $g$. The vertical displacement or height $h$ covered in time $t$ is given by the kinematic equation $h = ut + \frac{1}{2}gt^2$.
A stone dropped from the top of the tower touches the ground in $4\text{ sec}$. The height of the tower is about
A.
$80\text{ m}$
B.
$40\text{ m}$
C.
$20\text{ m}$
D.
$160\text{ m}$
Q38
DPT
DPT of Motion under Gravity
MCQ
12 Aug 2026
Concept: For a body falling freely from rest ($u = 0$) under gravity, the distance fallen in time $t$ is given by $s = \frac{1}{2}gt^2$. The separation between two bodies released at different times $t_1$ and $t_2$ from the same height is equal to the difference in their downward displacements: $s = s_1 - s_2 = \frac{1}{2}g(t_1^2 - t_2^2)$.
A body is released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies, two seconds after the release of the second body is
A.
$4.9\text{ m}$
B.
$9.8\text{ m}$
C.
$19.6\text{ m}$
D.
$24.5\text{ m}$