Question

A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.

Answer 1

Given the ball is dropped from the height of 90 m from the floor. The energy lost by the ball at every collision to the floor is one tenth of its speed.

Let u the initial velocity, v be the final velocity of the ball, h is the height from the ball is dropped and t is the time taken by the ball to reach at the ground.

The third equation of motion is,

v 2 − u 2 =2as

Substitute the required values in above equation.

v 2 − ( 0 ) 2 =2×9.81 m/ s 2 ×90 m v=42.02 m/s

The second equation of motion is,

s=ut+ 1 2 a t 2

Substitute the required values in above equation.

90 m=0×t+ 1 2 ×9.81 m/ s 2 × t 2 t=4.28 s

The velocity when the ball strikes the ground and moves in upward direction is,

v'= 9 10 ×v

Substitute the required values in above expression.

v ′ = 9 10 ×42.02 m/s =37.82 m/s

Let the time taken by the ball to reach at the maximum height be t ′ .

The first equation of motion is,

v= v ′ +a t ′

Substitute the required values in the above equation.

0=37.82 m/s +( −9.81 m/ s 2 ) t ′ t ′ =3.86 s

The total time taken in this round trip is,

T=t+ t ′

Substitute the required value in the above equation.

T=4.28 s+3.86 s =8.14 s

The velocity of the rebound is,

v ″ = 9 10 v ′

Substitute the required values in the above equation.

v ″ = 9 10 ×37.82 m/s =34.04 m/s

Now the total time of rebound is,

T ′ =T+ t ′

Substitute the required values in the above equation.

T ′ =8.14 s+3.86 s =12 s

The graph between the speed of the ball and time is as shown below,

The line OA represents the downward motion of the ball, line AB shows the upward motion of the ball.

The line BC shows the upward motion of the ball after striking the ground. Line CD represents the downward motion of the ball.

The line DE shows the upward motion of the ball.