A ball falls under gravity from a height 10 m with an initial velocity v_0. It hits the ground, loses 50% of its energy in collision and it rises to the same height, what is the value of v_0?

A ball falls under gravity from a height 10 m with an initial velocity

1.	A ball falls under gravity from a height 10 m with an initial velocity v_0. It hits the ground, loses 50% of its energy in collision and it rises to the same height, what is the value of v_0?
Answer» 14 m/s 7 m/s 28 m/s 9.8 m/s Solution :Let V be the velocity of the BALL when it hits the ground. then `v^2 = v_0^2 + 2gh = v_0^2 + 2 xx 9.8 xx 10` or `v^2 = v_0^2 + 196` Let `v.` be the velocity after impect and it RISE same height 10 m. `:. V.^(2) - 0 = 2 xx 9.8 xx 10 = 196 or v. = 14 m//s` Ratio of kinetic energy before impact and after impact `(1/2 mv^2)/(1/2 mv.^2) = (v^2)/(v.^2) = (v_0^2 + 196)/(196) = 2 " or " v_0 = 14 m//s` .