A ball is dropped from a height of 5 m. After it strikes the ground it bounces to a height of 2.45m.
If the impact with the ground lasts for 0.01 s, find average acceleration during the impact?
(g = 10 m/s)
Answers
Answered by
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
Explanation:
please mark me brilliant ❤️❤️
Similar questions