Question 12

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

Ball is dropped from a height, s = 90 m

Initial velocity of the ball, u = 0

Acceleration, a = g = 9.8 m/s2

Final velocity of the ball = v

From second equation of motion, time (t) taken by the ball to hit the ground can be obtained as:



s  =  ut  + ½ at2

90  =  0  + ½ x 9.8 x t2

t  =  underoot 18.38 =  4.29 s

From first equation of motion, final velocity is given as:

v = u + at

= 0 + 9.8 × 4.29 = 42.04 m/s

Rebound velocity of the ball, ur =  9/10 v  = 9/10 x 42.04 = 37.84 m/s

Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:

v = ur + at′ 

0 = 37.84 + (– 9.8) t′

t′  =  -37.84 / -9.8  = 3.86 s

Total time taken by the ball = t + t′ = 4.29 + 3.86 = 8.15 s

As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time.

The velocity with which the ball rebounds from the floor =  9/10 x 37.84  = 34.05 m/s

Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s

The speed-time graph of the ball is represented in the given figure as:

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