Question

of 10 m rebounds to a height of 2.5 m. If the
ball is in contact with the floor for 0.02s, its
A ball dropped on to the floor from a height
average acceleration during contact is
2) 1050ms-2
4) 9.8ms-2
1) 2100ms-2
3) 4200ms-2​

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