# 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.

Given,

Ball is dropped from a height = 90 m

Time interval, 0 ≤ t ≤ 12

Initial velocity of the ball =0 m/s

Final velocity of the ball = v m/s

Acceleration due to gravity, g = 9.81 ms-2

s = ut + 0.5gt2

where,

u = Initial velocity

g = Acceleration due to gravity

s = Distance covered

t = Time

90 = 0 + (0.5×9.81)t2

t = 4.29 s

v = u + gt

where,

v = Final velocity

u = Initial velocity

g = Acceleration due to gravity

t = Time

v = 0 + (9.81×4.29) = 42.04 m/s

Bounce velocity of the ball, vb = 0.9v = 37.84 m/s

Time (t’) by the bouncing ball to reach maximum is given by,

v = vb – gt’

where,

v = Final velocity

vb = Bounce velocity

g = Acceleration due to gravity

t’ = Bouncing time

0 = 37.84 – (9.81×t’)

t’ = 3.86 s

Total time taken by the ball = t + t’ = 4.29 + 3.86 = 8.15 s

The iterations go on like this up to ball reach a static condition.

The graph obtained by the data is,

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