Question

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​

Answer 1

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