A ball is dropped on the floor from a height of 80 m. It rebounds first time to a height of 20 m and second time to a height of 5 m. If the average acceleration of the ball is same during contact with floor, then find the ratio of time of contact during first bounce to second bounce ?
Answers
Answer:
2100m/s
2
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
The ratio of the time during first bounce to second bounce is 1 : 2
Ball is dropped from height = 80 m
In the first time, it is rebounded to a height = 20 m
So, the height difference (Δ h₁) = (80-20) m
⇒ Δ h₁ = 60 m
In the second time, it is rebounded to a height = 5 m
So, the height difference (Δ h₂) = (20-5) m
⇒ Δ h₂ = 15 m [As it is rebounded from the height of 20 m]
Now, we know, force, F = mΔv/t [m = mass of the object, t = time taken]
∵ Given, the average acceleration of the ball is same during contact with floor and the ball is same (so mass same),
∴ Δv ∝ 1/t
So, the ratio of the time,
t₁ : t₂ = Δv₂ : Δv₁
⇒ t₁ : t₂ = √(2gΔh₂) : √(2gΔh₁) [Δv = √(2gΔh), g = acceleration due to gravity]
⇒ t₁ : t₂ = √Δh₂ : √Δh₁
⇒ t₁ : t₂= √15 : √60
⇒ t₁ : t₂ = √15 : 2√15
⇒ t₁ : t₂ = 1 : 2
So, the ratio of time of contact during first bounce to second bounce is 1 : 2