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Question

When a rubber ball is dropped from a height of 2.5m onto a hard surface, it loses 25% of its mechanical energy on each bounce. With what speed would the ball have to be thrown downward to make it reach its original height after the first bounce?

Answer 1

So, what's gh/75%? Set that equal to the required kinetic energy+potential energy

Answer 2

♠ thus pot energy Ep = m*g*h on the roof is the same as kinetic Ek =0.5*m*v^2 on the ground;
♣ now bounce! and ΔE= 0.25*Ek= 0.25*Ep =0.25*m*g*h is lost and must be compensated by throwing the ball with such initial speed u that ΔE=0.5*m*u^2;
♦ therefore 0.5m*u^2 = 0.25mgh, hence u=sqrt(0.5*g*h) =3.5m/s;