A ball is dropped on a floor from a height of 5-meter.it rebounds to a height of 3.2 meters. if the ball is constant with the floor for 0.1 sec., then calculate its average acceleration during the contact, take g=10 m/s*s
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Answer:
Let u be the velocity with which the ball hits the ground, then
u
2
=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/se
Therefore A=Δt
Answer=Δv
=
0.01
21
=2100m/s
2