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# JEE Class main Answered

physics
Asked by aadityaghule2008 | 16 Dec, 2023, 10:57: PM
Expert Answer

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 vi 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 vi2 = m g h = ( 0.05 × 9.8 × 10 )  J = 4.9 J

Initial speed vi 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 vis calculated as

(1/2) m vf2 = 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)

Answered by Thiyagarajan K | 17 Dec, 2023, 08:42: AM
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