JEE Class main Answered

By conservation of energy, kinetic energy of ball just before making impact with ground  equals

its initial potential energy when the ball is at the height h = 10 m .

Let m = 0.05 kg be the mass of ball , g = 9.8 m/s² is acceleration due to gravity and v_i is speed of ball just before impact with ground

Hence , kinetic energy of ball just before making impact with ground is calculated as

(1/2) m v_i² = m g h = ( 0.05 × 9.8 × 10 )  J = 4.9 J

Initial speed v_i just before impact is calculated as

if 75%  of energy is lost after rebound , then final kinetic energy just afer impact  = (4.9/4) J = 1.225 J

Final speed v_f  is calculated as

(1/2) m v_f² = 1.225 J

Impulive force F from ground acting on the ball for a time duration Δt creates change in momentum Δ(mv)

F × Δt = Δ(mv)