Question

a body falling from a height of 10 m rebound from a hard floor, it loses 20% of its energy in impact. what is the height to which it rises after impact\

Answer 1

Height of the ball reached h1=10 metre
Let the potential energy be PE=mgh1       .....................eqn 1
And for the rebound let the height be h2
Energy loss is 20%
So the relation will be 
PE*20/100=mg(h1-h2)
now substituting eqn 1
we will get
⇒mgh1*20/100=mg(h1-h2)
⇒h1*20/100=h1-h2
⇒h2+h1*20/100=h1
⇒h2=h1- (h1*20/100)
⇒h2=h1{1-(20/100)}
⇒h2=10(80/100)
⇒h2=800/100
⇒h2 = 8 metres
Assumption : There is no friction between the air particles and the body.
Hence, the height reached after the impact is 8 metres.
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