Question

A ball dropped from a height of 10m, rebounds to a height of 2.5m. If the ball is in contact with the floor for 0.01 second, its acceleration during contact is (g = 9.8m/s?) 1) 20 m/s2 2) 21 m/s 3) 210 m/s2 no 4)2100 m/s ​

Answer 1

Explanation:

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

v

2

=2×9.8×2.5=49

⇒v=7m/sec

∴Δv=(v−u)

=(7m/sec)−(−14m/sec)

=21m/sec

∴a=

Δt

Δv

=

0.01

21

=2100m/s

2