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Motion in 1D

108 Questions Start DPT Test
All Exams Objective Physics Vol-1 DPT
Q51 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:
A.
40
B.
45
C.
48
D.
50
Q52 Objective Physics Vol-1 Basic Concepts of Motion MCQ
27 Jul 2026
Concept: Kinematic equations for uniformly accelerated motion
A car starts from rest, attains a velocity of $18 \mathrm{kmh}^{-1}$ with an acceleration of $0.5 \mathrm{ms}^{-2}$, travels 4 km with this uniform velocity and then comes to halt with a uniform deceleration of $0.4 \mathrm{ms}^{-2}$. Calculate the total time of travel of the car.
A.
812.5 s
B.
802.5 s
C.
822.5 s
D.
832.5 s
Q53 Objective Physics Vol-1 Basic Concepts of Motion MCQ
27 Jul 2026
Concept: Vector nature of acceleration and its dependence on velocity
Acceleration of a particle changes when
A.
direction of velocity changes
B.
magnitude of velocity changes
C.
Both (a) and (b)
D.
speed changes
Q54 Objective Physics Vol-1 Basic Concepts of Motion MCQ
27 Jul 2026
Concept: Relationship between acceleration, velocity, and speed
If a particle moves with an acceleration, then which of the following can remain constant?
A.
Both speed and velocity
B.
Neither speed nor velocity
C.
Only the velocity
D.
Only the speed
Q55 Objective Physics Vol-1 Basic Concepts of Motion MCQ
27 Jul 2026
Concept: Kinematic equations relating average velocity, distance, time, and acceleration
The average velocity of a body moving with uniform acceleration travelling a distance of $3.06 \mathrm{m}$ is $0.34 \mathrm{ms}^{-1}$. If the change in velocity of the body is $0.18 \mathrm{ms}^{-1}$, then during this time, its uniform acceleration is
A.
$0.01 \mathrm{ms}^{-2}$
B.
$0.02 \mathrm{ms}^{-2}$
C.
$0.03 \mathrm{ms}^{-2}$
D.
$0.04 \mathrm{ms}^{-2}$
Q56 Objective Physics Vol-1 Basic Concepts of Motion MCQ
27 Jul 2026
Concept: Calculation of retardation using the first equation of motion
A car travelling with a velocity of $80 \mathrm{km/h}$ slowed down to $44 \mathrm{km/h}$ in $15 \mathrm{s}$. The retardation is
A.
$0.67 \mathrm{ms}^{-2}$
B.
$1 \mathrm{ms}^{-2}$
C.
$1.25 \mathrm{ms}^{-2}$
D.
$1.5 \mathrm{ms}^{-2}$
Q57 Objective Physics Vol-1 Basic Concepts of Motion MCQ
27 Jul 2026
Concept: Acceleration in straight-line motion versus curved-path motion at constant speed
An object is moving along the path OABO with constant speed, then image.png
A.
the acceleration of the object while moving along the path OABO is zero
B.
the acceleration of the object along the path OA and BO is zero
C.
there must be some acceleration along the path AB
D.
Both (b) and (c)
Q58 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Kinematic equations for uniformly accelerated motion
Two cars start off a race with velocities $2 \mathrm{ms}^{-1}$ and $4 \mathrm{ms}^{-1}$ and travel in a straight line with uniform accelerations $2 \mathrm{ms}^{-2}$ and $1 \mathrm{ms}^{-2}$, respectively. What is the length of the path, if they reach the final point at the same time?
A.
12 m
B.
24 m
C.
36 m
D.
48 m
Q59 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Third equation of motion and calculation of retardation
A car was moving at a rate of $18 \mathrm{kmh}^{-1}$. When the brakes were applied, it comes to rest at a distance of 100 m. Calculate the retardation produced by the brakes.
A.
$0.125 \mathrm{ms}^{-2}$
B.
$0.250 \mathrm{ms}^{-2}$
C.
$0.500 \mathrm{ms}^{-2}$
D.
$1.000 \mathrm{ms}^{-2}$
Q60 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Stopping distance using the third equation of motion
Two cars are travelling towards each other on a straight road at velocities $10 \mathrm{ms}^{-1}$ and $12 \mathrm{ms}^{-1}$, respectively. When they are 150 m apart, both the drivers apply their brakes and each car decelerates at $2 \mathrm{ms}^{-2}$ until it stops. How far apart will they be when both of them come to rest?
A.
61 m
B.
75 m
C.
89 m
D.
100 m
Q61 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Relative motion and time of flight for constant velocity
A train travelling at $20 \mathrm{kmh}^{-1}$ is approaching a platform. A bird is sitting on a pole on the platform. When the train is at a distance of 2 km from the pole, brakes are applied which produce a uniform deceleration in it. At that instant, the bird flies towards the train at $60 \mathrm{kmh}^{-1}$ and after touching the nearest point on the train flies back to the pole and then flies towards the train and continues repeating itself. Calculate how much distance the bird covers before the train stops?
A.
6 km
B.
12 km
C.
18 km
D.
24 km
Q62 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Kinematic equations for uniformly accelerated motion applied in segments
A particle starts with an initial velocity and passes successively over the two halves of a given distance with accelerations $a_1$ and $a_2$, respectively. The final velocity is the same as if the whole distance is covered with a uniform acceleration of
A.
$\frac{a_1 - a_2}{2}$
B.
$\frac{2a_1 a_2}{a_1 + a_2}$
C.
$\frac{a_1 + a_2}{2}$
D.
$\sqrt{a_1 a_2}$
Q63 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Equations of motion for two bodies starting from rest
In a car race, car A takes a time $t$ less than car B at the finish point and passes the finishing point with speed $v$ more than that of the car B. Assuming that both the cars start from rest and travel with constant accelerations $a_1$ and $a_2$ respectively, the correct relation is
A.
$v = \sqrt{a_1 a_2}t$
B.
$v = (a_1 + a_2)t$
C.
$v = \sqrt{\frac{a_1}{a_2}}t$
D.
$v = \frac{a_1 a_2}{t}$
Q64 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Distance travelled in the nth second of uniformly accelerated motion
A body starting from rest has an acceleration of $4 \mathrm{ms}^{-2}$. The distance travelled by it in the 5th second is
A.
20 m
B.
18 m
C.
16 m
D.
24 m
Q65 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Ratio of displacements in successive time intervals for a body starting from rest
A particle starts from rest and moves under constant acceleration in a straight line. The ratio of displacements in successive seconds is
A.
1:2:3:...
B.
1:4:9:...
C.
2:4:6:...
D.
1:3:5:...
Q66 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: First equation of motion for uniformly accelerated motion
Velocity of a body moving along a straight line with uniform acceleration reduces by $\frac{3}{4}$ of its initial velocity in time $t_{0}$. The total time of motion of the body till its velocity becomes zero is
A.
$\frac{4}{3}t_{0}$
B.
$\frac{3}{2}t_{0}$
C.
$\frac{5}{3}t_{0}$
D.
$\frac{8}{3}t_{0}$
Q67 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Second equation of motion for displacement
The displacement of a body in 8 s starting from rest with an acceleration of $20 \mathrm{cms}^{-2}$ is
A.
$64 \mathrm{m}$
B.
$64 \mathrm{cm}$
C.
$640 \mathrm{cm}$
D.
$0.064 \mathrm{m}$
Q68 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Integration of velocity to find displacement and distance
The motion of a particle is described by the equation $v = at$. The distance travelled by the particle in the first 4 s is
A.
$4a$
B.
$12a$
C.
$6a$
D.
$8a$
Q69 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Second equation of motion for calculating time
A particle starts with a velocity of $2 \mathrm{ms}^{-1}$ and moves in a straight line with a retardation of $0.1 \mathrm{ms}^{-2}$. The first time at which the particle is 15 m from the starting point is
A.
10 s
B.
20 s
C.
30 s
D.
40 s
Q70 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Calculation of total distance in different phases of motion
A particle starts from rest, accelerates at $2 \mathrm{ms}^{-2}$ for 10 s and then moves with constant speed of $20 \mathrm{ms}^{-1}$ for 30 s and then decelerates at $4 \mathrm{ms}^{-2}$ till it stops after next 5 s. What is the distance travelled by it?
A.
750 m
B.
800 m
C.
700 m
D.
850 m
Q71 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Equation of motion for uniform velocity and uniform acceleration
A body is moving with uniform velocity of $8 \mathrm{ms}^{-1}$. When the body just crossed another body, the second one starts and moves with uniform acceleration of $4 \mathrm{ms}^{-2}$. The time after which two bodies meet, will be
A.
2 s
B.
4 s
C.
6 s
D.
8 s
Q72 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Second equation of motion for displacement
Two bodies A and B start from rest from the same point with a uniform acceleration of $2 \mathrm{ms}^{-2}$. If B starts one second later, then the two bodies are separated at the end of the next second, by
A.
1 m
B.
2 m
C.
3 m
D.
4 m
Q73 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Calculation of average velocity using total displacement and total time
A train accelerating uniformly from rest attains a maximum speed of $40 \mathrm{ms}^{-1}$ in 20 s. It travels at this speed for 20 s and is brought to rest by uniform retardation in further 40 s. What is the average velocity during this period?
A.
$\frac{80}{3} \mathrm{ms}^{-1}$
B.
$40 \mathrm{ms}^{-1}$
C.
$25 \mathrm{ms}^{-1}$
D.
$30 \mathrm{ms}^{-1}$
Q74 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Relation between average speed, maximum speed, and distances in different phases of motion
A particle starts from rest and traverses a distance $l$ with uniform acceleration, then moves uniformly over a further distance $2l$ and finally comes to rest after moving a further distance $3l$ under uniform retardation. Assuming entire motion to be rectilinear motion, the ratio of average speed over the journey to the maximum speed on its ways is
A.
1/5
B.
2/5
C.
3/5
D.
4/5
Q75 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Third equation of motion applied to find velocity at the midpoint
A body travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $20 \mathrm{ms}^{-1}$ and $30 \mathrm{ms}^{-1}$, respectively. The speed of the body at mid-point of $A$ and $B$ is
A.
$25 \mathrm{ms}^{-1}$
B.
$25.5 \mathrm{ms}^{-1}$
C.
$24 \mathrm{ms}^{-1}$
D.
$10\sqrt{6} \mathrm{ms}^{-1}$
Q76 Objective Physics Vol-1 Uniform Accelerated MCQ
27 Jul 2026
Concept: Distance travelled in the nth second of uniformly accelerated motion
If a body starts from rest and travels 120 cm in the 6th second, then what is the acceleration?
A.
$0.20 \mathrm{ms}^{-2}$
B.
$0.027 \mathrm{ms}^{-2}$
C.
$0.218 \mathrm{ms}^{-2}$
D.
$0.03 \mathrm{ms}^{-2}$
Q77 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Kinematic equations for motion under gravity
A ball is thrown upwards from the top of a tower 40 m high with a velocity of 10 m/s. Find the time when it strikes the ground. (Take, $g = 10 \mathrm{m/s}^{2}$)
A.
4 s
B.
2 s
C.
5 s
D.
3 s
Q78 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity in two phases
A rocket is fired vertically up from the ground with a resultant vertical acceleration of $10 \mathrm{ms}^{-2}$. The fuel is finished in 1 min and it continues to move up. What is the maximum height reached by the rocket? (Take, $g = 10 \mathrm{ms}^{-2}$)
A.
18 km
B.
54 km
C.
36 km
D.
72 km
Q79 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity and time of ascent
A juggler throws balls into air. He throws one ball whenever the previous one is at its highest point. How high does the balls rise, if he throws $n$ balls each second? (Acceleration due to gravity is $g$)
A.
$\frac{g}{n^{2}}$
B.
$\frac{g}{2n^{2}}$
C.
$\frac{g}{4n^{2}}$
D.
$\frac{2g}{n^{2}}$
Q80 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Kinematic equations for motion under gravity
From an elevated point A, a stone is projected vertically upwards. When the stone reaches a distance $h$ below A, its velocity is double of what it was at a height $h$ above A. What is the greatest height attained by the stone?
A.
$\frac{h}{3}$
B.
$\frac{2h}{3}$
C.
$\frac{4h}{3}$
D.
$\frac{5h}{3}$
Q81 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Kinematic equations for motion under gravity
A ball is thrown vertically upwards with a velocity of $20 \mathrm{ms}^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is 25 m from the ground. How long will it take before the ball hits the ground? (Take, $g = 10 \mathrm{ms}^{-2}$)
A.
5 s
B.
2 s
C.
3 s
D.
4 s
Q82 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Kinematic equations for motion under gravity and quadratic equation
A ball is thrown upwards from the ground with an initial speed $u$. The ball is at a height of 80 m at two times, for the time interval of 6 s. Find the value of $u$. (Take, $g = 10 \mathrm{ms}^{-2}$)
A.
$30 \mathrm{ms}^{-1}$
B.
$40 \mathrm{ms}^{-1}$
C.
$50 \mathrm{ms}^{-1}$
D.
$60 \mathrm{ms}^{-1}$
Q83 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Time of flight and kinematic equations for motion under gravity
A particle is thrown vertically upwards from the surface of the earth. Let $T_{P}$ be the time taken by the particle to travel from a point P above the earth to its highest point and back to the point P. Similarly, let $T_{Q}$ be the time taken by the particle to travel from another point Q above the earth to its highest point and back to the same point Q. If the distance between the points P and Q is $H$, find the expression for acceleration due to gravity in terms of $T_{P}$, $T_{Q}$ and $H$.
A.
$g = \frac{4H}{T_{P}^{2} - T_{Q}^{2}}$
B.
$g = \frac{8H}{T_{P}^{2} - T_{Q}^{2}}$
C.
$g = \frac{8H}{T_{P}^{2} + T_{Q}^{2}}$
D.
$g = \frac{2H}{T_{P}^{2} - T_{Q}^{2}}$
Q84 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Free fall under gravity at equal time intervals
From the top of a building 16 m high, water drops are falling at equal intervals of time such that when the first drop reaches the ground, the fifth drop just starts. Find the distance between the successive drops at that instant.
A.
7 m, 5 m, 3 m, 1 m
B.
5 m, 4 m, 3 m, 2 m
C.
6 m, 4 m, 3 m, 1 m
D.
8 m, 5 m, 3 m, 2 m
Q85 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Relative motion of two freely falling bodies
A ball is dropped from the top of a tower. After 2 s, another ball is thrown vertically downwards with a speed of $40 \mathrm{ms}^{-1}$. After how much time and at what distance below the top of the tower do the balls meet? (Take, $g = 10 \mathrm{ms}^{-2}$)
A.
2 s, 20 m
B.
3 s, 45 m
C.
4 s, 80 m
D.
5 s, 125 m
Q86 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Non-uniformly accelerated motion and calculus-based kinematics. When acceleration is not constant, the basic equations of velocity and acceleration are derived using differentiation and integration. Velocity is the rate of change of displacement ($v = \frac{ds}{dt}$), and acceleration is the rate of change of velocity ($a = \frac{dv}{dt}$). Displacement can be found by integrating the velocity equation with respect to time, applying the given boundary conditions.
The velocity-time equation of a particle moving in a straight line is given by $v = 10 + 2t + 3t^2$ (in SI units). If the displacement of the particle is 20 m at time $t = 0$, what is the displacement of the particle at time $t = 1$ s and its corresponding acceleration-time equation?
A.
Displacement = 32 m, Acceleration $a = 2 + 6t$
B.
Displacement = 20 m, Acceleration $a = 2 + 3t$
C.
Displacement = 32 m, Acceleration $a = 10 + 2t$
D.
Displacement = 12 m, Acceleration $a = 6t$
Q87 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Non-uniformly accelerated motion and calculus-based kinematics. When velocity is given as a function of displacement ($v = f(x)$), we can use the relation $v = \frac{dx}{dt}$ to separate variables and integrate to find the displacement-time or velocity-time relationship. The mean velocity is defined as the total displacement divided by the total time taken.
The velocity of a particle moving in the positive direction of the X-axis varies as $v = \alpha \sqrt{x}$, where $\alpha$ is a positive constant. Assuming that at moment $t = 0$, the particle was located at the point $x = 0$. What is the time dependence of the velocity of the particle and the mean velocity of the particle averaged over the time that the particle takes to cover the first $s$ metres of the path?
A.
$v = \alpha^2 t$, Mean velocity = $\alpha \sqrt{s}$
B.
$v = \frac{1}{2} \alpha^2 t$, Mean velocity = $\frac{\alpha \sqrt{s}}{2}$
C.
$v = \frac{1}{2} \alpha^2 t$, Mean velocity = $\alpha \sqrt{s}$
D.
$v = \alpha^2 t$, Mean velocity = $\frac{\alpha \sqrt{s}}{2}$
Q88 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity. When an object is thrown vertically upwards, the time of ascent is equal to the time of descent. The total time of flight is given by $T = \frac{2u}{g}$.
If a stone is thrown up with a velocity of $9.8 \text{ ms}^{-1}$, then how much time will it take to come back?
A.
1 s
B.
2 s
C.
3 s
D.
4 s
Q89 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity. The distance covered in the last $t$ seconds of ascent is equal to the distance covered in the first $t$ seconds of free fall from the maximum height, where the initial velocity is zero.
If a ball is thrown vertically upwards with speed $u$, the distance covered during the last $t$ second of its ascent is
A.
$ut - (gt^2 / 2)$
B.
$(u + gt)t$
C.
$ut$
D.
$gt^2 / 2$
Q90 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity. The time taken to reach the maximum height is the time of ascent. If the next ball is thrown when the first ball's velocity is zero, the time of ascent is equal to the interval between throws. The maximum height can be found using $h = \frac{1}{2}gt^2$.
A person throws balls into air after every second. The next ball is thrown when the velocity of the first ball is zero. How high do the balls rise above his hand?
A.
2 m
B.
5 m
C.
8 m
D.
10 m
Q91 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity. Using the third equation of motion $v^2 - u^2 = 2as$, we can relate the velocities at different heights. At maximum height, the final velocity is zero.
A particle is thrown vertically upwards. Its velocity at half of the height is $10 \text{ ms}^{-1}$. Then, the maximum height attained by it is (Take, $g = 10 \text{ ms}^{-2}$)
A.
16 m
B.
10 m
C.
20 m
D.
40 m
Q92 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity. The maximum height attained by a body thrown vertically upwards is directly proportional to the square of its initial velocity, given by $h = \frac{v_0^2}{2g}$.
When a ball is thrown up vertically with velocity $v_0$, it reaches a maximum height of $h$. If one wishes to triple the maximum height, then the ball should be thrown with velocity,
A.
$\sqrt{3}v_0$
B.
$3v_0$
C.
$9v_0$
D.
$3/2v_0$
Q93 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity. Using the third equation of motion $v^2 - u^2 = 2as$, taking the downward direction as positive to relate the initial velocity, final velocity, and the height of the tower.
A stone thrown upward with a speed $u$ from the top of the tower reaches the ground with a speed $3u$. The height of the tower is
A.
$3u^2 / g$
B.
$4u^2 / g$
C.
$6u^2 / g$
D.
$9u^2 / g$
Q94 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity. The time interval between passing a certain height twice is the time it takes to go from that height to the maximum height and back. The velocity at that height can be found using the time of flight for that specific segment, assuming $g = 10 \text{ ms}^{-2}$ for calculation.
A body thrown vertically up from the ground passes the height of 10.2 m twice in an interval of 10 s. What was its initial velocity?
A.
$52 \text{ ms}^{-1}$
B.
$61 \text{ ms}^{-1}$
C.
$45 \text{ ms}^{-1}$
D.
$26 \text{ ms}^{-1}$
Q95 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity. The total time of flight to reach a certain height and return is $\frac{2u}{g}$. The time to reach the point is $t_1$, and the time to return from the peak is determined by the symmetry of the motion.
A body is projected with a velocity $u$. It passes through a certain point above the ground after $t_1$ second. The time interval after which the body passes through the same point during the return journey is
A.
$\left(\frac{u}{g} - t_1^2\right)$
B.
$2\left(\frac{u}{g} - t_1\right)$
C.
$\left(\frac{u}{g} - t_1\right)$
D.
$\left(\frac{u^2}{g^2} - t_1\right)$
Q96 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity. Using the second equation of motion $s = ut + \frac{1}{2}at^2$ for the three different cases (upward, downward, and free fall) from the same height to find the relationship between the times.
A body is thrown vertically upwards from the top A of tower. It reaches the ground in $t_1$ second. If it is thrown vertically downwards from A with the same speed, it reaches the ground in $t_2$ second. If it is allowed to fall freely from A, then the time it takes to reach the ground is given by
A.
$t = \frac{t_1 + t_2}{2}$
B.
$t = \frac{t_1 - t_2}{2}$
C.
$t = \sqrt{t_1 t_2}$
D.
$t = \sqrt{\frac{t_1}{t_2}}$
Q97 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity and relative motion. First, find the velocity and height of the balloon when the ball is released. Then, treat the ball as a projectile thrown upwards with that initial velocity from that height, under the influence of gravity.
A man in a balloon rising vertically with an acceleration of $4.9 \text{ ms}^{-2}$ releases a ball 2 s after the balloon is let go from the ground. The greatest height above the ground reached by the ball is (Take, $g = 9.8 \text{ ms}^{-2}$)
A.
14.7 m
B.
19.6 m
C.
9.8 m
D.
24.5 m
Q98 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity. The distance covered in the $n$-th second of a freely falling body is $S_n = \frac{1}{2}g(2n - 1)$, and the distance covered in the first $t$ seconds is $S = \frac{1}{2}gt^2$.
A stone falls freely under gravity. The total distance covered by it in the last second of its journey equals the distance covered by it in first 3 s of its motion. The time for which stone remains in air, is
A.
5 s
B.
12 s
C.
15 s
D.
8 s
Q99 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity. The distance travelled in time $t$ is $S = \frac{1}{2}gt^2$. We can find the distance travelled in successive 2-second intervals and then determine their ratio.
A body falls from a height $h = 200 \text{ m}$. The ratio of distance travelled in each $2 \text{ s}$, during $t = 0$ to $t = 6 \text{ s}$ of the journey is
A.
1:4:9
B.
1:2:4
C.
1:3:5
D.
1:2:3
Q100 Objective Physics Vol-1 Motion under Gravity MCQ
27 Jul 2026
Concept: Motion under gravity. The time taken by a body to fall freely from a height $H$ is given by $t = \sqrt{\frac{2H}{g}}$.
A ball is released from height $h$ and another from $2h$. The ratio of time taken by the two balls to reach the ground is
A.
$1:\sqrt{2}$
B.
$\sqrt{2}:1$
C.
2:1
D.
1:2