It is given that Ball is dropped from a height = 90 m
Time interval, 0 to 12 sec
Initial velocity of the ball , u =0 m/s
Acceleration due to gravity , g = 9.8 m/s²
s = ut + 1/2 gt2
where, u = Initial velocity
g = Acceleration due to gravity
s = Distance covered
t = Time
⇒ 90 = 0 + 1/2 × 9.8 × t²
⇒ 90 = 4.9t²
⇒ t² = 900/49
∴ t = 30/7 = 4.28 s
From 1st equation of motion for freely falling body,
v = u + gt
Where u = 0, g = 9.8 m/s² and t = 4.28s
⇒ v = 0 + (9.8 ×4.28) ≈ 42 m/s
Bounce velocity of the ball, vb = 0.9v = 37.8 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 – (9.8 ×t’)
∴ t’ = 3.86 s
Total time taken by the ball = t + t’ = 4.29 + 3.86 = 8.15 s
The will continue like this up to ball reach a static condition.
The graph obtained by the data is,