# Applications of equations of motion

## Trending Questions

**Q.**A scooter accelerates from rest for time t1 at a constant rate α1 and then retards at constant rate α2 for time t2 and comes to rest. The correct value of t1t2 will be -

- α1+α2α2
- α2α1
- α1+α2α1
- α1α2

**Q.**A ball is thrown vertically downwards with speed 30 m/s from the top of the tower of height 80 m. The speed of the ball with which it hits the grounds is (g=10 m/s2)

- 40 m/s
- 60 m/s
- 50 m/s
- 30 m/s

**Q.**A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is :

(g=10 m/s2)

- 340 m
- 360 m
- 300 m
- 320 m

**Q.**A stone is released from the top of the tower of height 20 m. Then its final velocity just before touching the ground will be

- 20 m/s
- 40 m/s
- 200 m/s
- 400 m/s

**Q.**A car starts from rest and accelerates uniformly to a speed of 180 km/h in 10 s. The distance covered by the car in this time interval is

- 500 m
- 250 m
- 100 m
- 200 m

**Q.**A man is d distance behind a bus. The bus starts from rest and moves away from the man with an acceleration a. At the same time man starts running towards the bus with a constant velocity v. Which of the following options are correct?

- the man catches the bus if v≥√2 ad
- if man just catches the bus, the time of catching the bus will be t=v/a
- if man just catches the bus, the time of catching the bus will be t=2v/a
- the man will catch the bus if v≥√ad

**Q.**A person throws vertically 10 balls per second with the same velocity. He throws a ball whenever the previous one is at its highest point. The height to which the balls rise is

- 10 m
- 5 m
- 5 cm
- 10 cm

**Q.**A ball is thrown vertically upwards with speed 10 ms−1 from the top of a tower. It hits the ground after some time with a speed of 50 ms−1. The height of tower is (g=10 ms−2)

- 240 m
- 120 m
- 60 m
- 80 m

**Q.**A body is projected vertically upward with speed 40 m/s. The distance travelled by the body in last 2 second of the journey is [g=10 m/s2]

- 10 m
- 5 m
- 60 m
- 15 m

**Q.**The acceleration of the particle moving in x−direction with an initial velocity 8 m/s is 3 m/s2. Find its velocity at t=2 s.

- 14 m/s
- 20 m/s
- 18 m/s
- 60 m/s

**Q.**A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2ms2. He reaches the ground with a speed of 3ms. At what height, did he bail out ?

- 293 m
- 111 m
- 91 m
- 182 m

**Q.**A truck has an initial velocity of 50 m/s. On the application of brakes truck stops at 200 m. The acceleration of the truck will be

- −12.5 ms−2
- 24.5 ms−2
- −6.25 ms−2
- 6.25 ms−2

**Q.**A ball thrown vertically upwards with a speed of 19.6 ms−1 from the top of a tower returns to the earth in 6 s. Find the height of the tower.

(Take g=9.8 m/s2)

- 27.8 m
- 58.8 m
- 70 m
- 92 m

**Q.**A body is thrown vertically upwards with velocity u. The distance travelled by it in the fifth and the sixth seconds are equal. The velocity u is given by (g=9.8 m/s2)

- 24.5 m/s
- 49.0 m/s
- 73.5 m/s
- 98.0 m/s

**Q.**A particle starting from rest travels a distance x in first 2 seconds and a distance y in next two seconds, then

- y=x
- y=2x
- y=3x
- y=4x

**Q.**Two bodies A (of mass 1 kg) and B (of mass 3 kg) are dropped from heights of 16 m and 25 m, respectively. The ratio of the time taken by them to reach the ground is

- 45
- 54
- 125
- 512

**Q.**Assume that a car is able to stop with a retardation of 8 m/s2 and that a driver can react to an emergency in 0.5 sec. Calculate the overall stopping distance of the car for a speed of 72 km/hr of the car

- 25 m
- 10 m
- 35 m
- 30 m

**Q.**A ball is thrown vertically downward with a velocity u from the top of a tower. It strikes the ground with a velocity 5u. The time taken by the ball to reach the ground is given by

- 2.5u
- 2ug
- 4ug
- 5u

**Q.**A stone dropped from a certain height reaches ground in 5 sec. If it is dropped and stopped for infinitesimal time interval at t=3 s, then time taken by the stone to reach the ground is

- 6 s
- 6.5 s
- 7 s
- 7.5 s

**Q.**The acceleration of the particle moving in positive x−direction with an initial velocity of 10 m/s is 3 m/s2. Its velocity at t=4 sec is

- 20 m/s
- 22 m/s
- 24 m/s
- 18 m/s

**Q.**A car has an initial velocity of 60 m/s. On application of brakes the car stops after 300 m. The retardation (deceleration) of the car is

- 5 m/s2
- 3 m/s2
- 6 m/s2
- 10 m/s2

**Q.**The brakes applied to a bike produce retardation of 5 m/s2. If the bike takes 4 s to stop after the application of brakes, calculate the distance it travels during this time.

- 40 m
- 45 m
- 35 m
- 50 m

**Q.**A ball is thrown vertically upwards with velocity of 10 m/s from the top of a tower. It returns to ground after some time with speed of 60 m/s. The height of the tower is

(g=10 m/s2)

- 375 m
- 175 m
- 125 m
- 225 m

**Q.**A sports car can accelerate uniformaly to a speed of 162 km/h in 5 sec. Its maximum breaking retardation is 6 m/s2. Minimum time in which it can travel 1 km starting from rest and ending at rest in seconds is?

**Q.**A ball is thrown vertically upwards from the top of a tower at 4.9 m/s. It strikes the pond near the base of the tower after 3 seconds . The height of the tower is

- 73.5 m
- 44.1 m
- 29.4 m
- None of these

**Q.**A ball at rest is dropped into a well. The water is at a depth h from the surface. If the speed of sound is c, then the time after which the splash is heard will be

- h[√2gh+1c]
- h[√2gh−1c]
- h[2g+1c]
- h[2g−1c]

**Q.**An open lift is moving upwards with velocity 10 m/s. It has an upward acceleration of 2 m/s2. A ball is projected upwards with velocity 20 m/s relative to ground. Find the time when ball again meets the lift.

- 43 s
- 53 s
- 54 s
- None

**Q.**

A ball is in a hot air balloon that starts from rest at t = 0 and accelerates upwards at 2m/s^{2}. After 10s, the ball is dropped from the balloon.Draw ball's v - t graph:

**Q.**A balloon is moving vertically upward with a velocity of 5 m/s. When it is at a height of h, a body is gently released from it. If it reaches the ground in 3 s, the height of the balloon when the body is released is (Take g=10 m/s2)

- 45 m
- 60 m
- 30 m
- 15 m

**Q.**You drop a ball from a window located on an upper floor of a building. It strikes the ground with a speed V. You now repeat the drop, but your friend down on the ground throws another ball upward at the same speed V, releasing his ball at the same moment at which you drop yours from the window. At some location, the balls pass each other. This location is:

- At the halfway point between window and ground
- Above halfway point
- Below halfway point
- Cannot be determined