Question

A ball is dropped on the floor from a height of 10m
At rebounds to a height of 2.5 m. if the ball is in
contact with the floor fox o.01 sec, the
average
acceleration during contact is​

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

Answer 2

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

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