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
From 2nd equation of motion for freely falling body,
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
From 1st equation of motion for freely falling body,
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,
