A ball is drop from a height of 20m and rebounce with a velocity which is 3/4 of this velocity with which it hit the ground. What is the time interval between the first and second bounces? (g=9.8m/s2)
one needs to calculate the velocity of the bounce:
1/2 m v^2=mgh
v= sqrt 2gh
vrebound=3/4 sqrt 40g
Now, figure the time in air for that ball:
hf=hi+vi*t-4.8t^2
0=0+3/4 sqrt(40g) t-4.8t^2
t=3sqrt(40g)/(4*4.8) seconds
16.8mm
16.8seconds
To find the time interval between the first and second bounces of a ball, we need to calculate the time it takes for the ball to hit the ground initially and then the time it takes for the ball to rebound back to its original height.
Let's break down the problem step by step:
1. First, calculate the time it takes for the ball to hit the ground initially when dropped from a height of 20m.
The initial height (h0) = 20m
Acceleration due to gravity (g) = 9.8m/s^2
We can use the kinematic equation for free fall:
h = h0 + (v0)t + (1/2)gt^2
Since the ball is dropped from rest, the initial velocity (v0) is 0.
Plugging in the values:
0 = 20 + (0)t + (1/2)(9.8)t^2
This equation simplifies to:
4.9t^2 = 20
To solve for t, divide both sides by 4.9:
t^2 = 4.08
Taking the square root of both sides:
t ≈ 2.02 seconds
Therefore, it takes approximately 2.02 seconds for the ball to hit the ground initially.
2. Next, calculate the time it takes for the ball to rebound back to its original height.
The velocity with which the ball rebounds is 3/4 of the velocity with which it hit the ground. Let's call this velocity vr.
vr = (3/4)v0
To find vr, let's use the equation of motion:
v = v0 + gt
Plugging in the values:
vr = (3/4)(0) ≈ 0
Since the velocity is approximately 0 when the ball reaches its maximum height, the time it takes for the ball to rebound back to its original height is the same as the time it takes for the ball to reach its maximum height after the first bounce.
So, the time interval between the first and second bounces is approximately 2.02 seconds.