Asked by Jeevan

A solid rubber ball has a diameter of 8.0 cm. It is released from rest with the top of the ball 80 cm above a horizontal surface. It falls vertically and then bounces back up so that the maximum height reached by the top of the ball is 45 cm, as shown.
80 60 40 20
If the kinetic energy of the ball is 0.75 J just before it strikes the surface, what is its kinetic energy just after it leaves the surface?
Answered by MathMate
Unless I misunderstood the numbers, the value 45 cm does not match the sequence 80,60,40,20.
Please check or clarify.
Answered by Nmk
the 45cm is supposed to be between 60cm to 40cm. The question is correct. Anyone who can help me in this too?
Answered by may
It says top of the ball, so you need to minus both of the heights by 8cm.
h1 = 0.8 -0.08 = 0.72
It is given that kinetic energy before striking the surface is 0.75 J. So you just nid to use potential energy formula(mgh) to calculate the mass of the ball .

h2 = 0.45 - 0.08 = 0.37
Use the mass that you had calculate and formula mgh again to count the kinetic energy after it leaves the surface.
Answered by GeeksWork
.45-.04=0.41
.80-.04=.76
(.41/.76)×0.75
=0.39
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