A soft tennis ball is dropped onto a hard floor from a height of 1.60 m and rebo

Question

A soft tennis ball is dropped onto a hard floor from a height of 1.60 m and rebounds to a height of 1.19 m. (Assume that the positive direction is upward.)

(a) Calculate its velocity just before it strikes the floor.
m/s
(b) Calculate its velocity just after it leaves the floor on its way back up.
m/s
(c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms.
m/s2
(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
m

Explanation / Answer

a)

Vi = velocity just before hiting the floor

hi = initial height from which the ball was dropped = 1.60 m

using conservation of energy

kinetic energy at bottom = potential energy

(0.5) m Vi2 = m ghi

Vi = sqrt(2g hi) = sqrt(2 x 9.8 x 1.60) = 5.6 m/s

since velocity is in downward direction which is taken as negative , hence

Vi = - 5.6 m/s

b)

Vf = final velocity just after the collision with the floor in up direction

hf = final height gained after collision = 1.19

using conservation of energy

kinetic energy at bottom = potential energy

(0.5) m Vf2 = m ghf

Vf = sqrt(2g hf) = sqrt(2 x 9.8 x 1.19) = 4.83 m/s

c)

t = time of contact = 0.0035 sec

acceleration is given as

a = (Vf - Vi)/t = (4.83 - (-5.6))/0.0035 = 2980 m/s2

d)

consider the motion during collision

Vi = initial velocity = 5.6 m/s

Vf = final velocity = 0 m/s

a = acceleration = - 2980

d = compression of ball

Using the equation

Vf2 = Vi2 + 2 a d

02 = 5.62 + 2 (-2980) d

d = 0.00526 m

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