Applying Newton's Third Law of Mortion

[Rosieposy08]

Homework Statement

After falling from a rest from a height of 30 m, a .50 kg ball rebounds upward, reaching a height of 20 m. If the contact between ball and ground lasted 2.0 ms, what average force was exerted on the ball?

Homework Equations

2.0ms=.002s

The Attempt at a Solution

 

[leachlife4]

Consider the total energy and direction (and change thereof) the ball has before and after the impact.

 

[supratim1]

hello Rosieposy! Welcome to physics forums!

you need to show your attempt at solving the question, if you are stuck people here will help you.

 

[ashishsinghal]

Use impulse = change in momentum.
BTW how is this an application of Newton's Third Law of Motion. As I recall it was action reaction law. You are not considering the force exerted on the ground by the ball, are you?

 

[rwisz]

Using Newton's Third Law of Motion, we know that for every action, there is an equal and opposite reaction. In this case, the action is the ball hitting the ground and the reaction is the ground exerting a force on the ball. Therefore, the average force exerted on the ball can be calculated by using the formula F=ma, where F is the force, m is the mass of the ball, and a is the acceleration.

First, we need to find the acceleration of the ball. We know that the ball fell from a height of 30 m and rebounded to a height of 20 m, so the change in height is 30 m - 20 m = 10 m. Using the formula vf^2 = vi^2 + 2ad, we can find the final velocity of the ball when it reached a height of 20 m:

vf^2 = 0 + 2(-9.8 m/s^2)(10 m)
vf = √(196 m^2/s^2)
vf = 14 m/s

Now, using the formula a = (vf - vi)/t, we can find the acceleration of the ball during the contact with the ground:

a = (14 m/s - 0)/0.002 s
a = 7000 m/s^2

Finally, using F=ma, we can calculate the average force exerted on the ball:

F = (0.50 kg)(7000 m/s^2)
F = 3500 N

Therefore, the average force exerted on the ball during the 2.0 ms contact with the ground was 3500 Newtons.