Question

A ball is dropped from a height of 10m. If the energy of the ball  reduces by 40percent after striking ground, how high ball bounce

Answer 1

H₁ =10 m, v₁ =?

Loss of energy =40% H₂ =?

On striking the floor, K.E = P.E

½ mv₁² = mg H₁

V₁ = ?(2gH₁)= ?( 2x9.8x10)

V₁= ?196= 14 m/s

As the body loses 40% of its energy, final energy E₂ =( 60/100) of initial energy

i.e. mgH₂ = (60/100)mgH₁

H₂ =3/5 H₁= (3/5)X10= 6m