Question

Marks: 1/20
A ball is dropped on to the floor from
a height of 20 m. It rebounds to a
height of 10 m. If the ball is in
contact with the floor for 0.1
seconds, what is the
average acceleration during contact?
[Take g = 10ms-2​

Answer 1

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

Explanation:

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