A ball dropped on to the floor from a height of 10m rebounds to a height of 2.5m. If the ball is in
contact with the floor for 0.02s, its averge acceleration during contact is
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A ball dropped on to the floor from a height of 10m rebounds to a height of 2.5m. If the ball is in
contact with the floor for 0.02s, its averge acceleration during contact is
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Explanation:
Let u be the velocity with which the ball hits the ground, then
u^2=2gh
u^2=2×9.8×10=196
u=14m/sec
If v be the velocity with which it rebounds, then
v^2=2×9.8×2.5=49
v=7m/sec
t=0.01s
•a=(v−u)/t
a=(7m/sec)−(−14m/sec)/0.01s
a=2100m/s^2