A ball is dropped on the floor from a height of 10m. it rebounds to a height of 2.5m. if the ball is in contact with ground for 0.02secs its average velocity is
1) 2100 ms^-2
2) 1050ms^-2
3) 4200ms^-2
4) 9.8ms^-2
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2100m/s2
Let u be the velocity with which the ball hits the ground, then
u2=2gh
=2×9.8×10=196
∴u=14m/sec
If v be the velocity with which it rebounds, then
v2=2×9.8×2.5=49
⇒v=7m/sec
∴Δv=(v−u)
=(7m/sec)−(−14m/sec)
=21m/sec
∴a=ΔtΔv=0.0121
=2100m/s2
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