A ball with a mass of 2.89 kg bounces off of the ground. Just before it hits the ground, its velocity is 6.00 m/s in the downward direction. Right after it bounces, its velocity is 4.54 m/s in the upward direction. What is the magnitude of the impulse J that the ground gave the ball? Give your answer in kg m/s with an accuracy of 0.1 kg m/s.

30.46 kgm/s

Explanation:

According to conservation law of momentum, the magnitude of the impulse J that the ground gave the ball equals to the change in momentum of the ball before and after it hits the ground.

Before the hit, the ball velocity is 6m/s, so its momentum is 6 * 2.89 = 17.34 kgm/s

After the hit, the ball velocity is -4.54 m/s in the opposite direction, so its momentum is 2.89*(-4.54) = -13.12 kgm/s

So the change of momentum, and also the impulse is

17.34 - (-13.12) = 30.46 kgm/s

1/0.14nm = 10^10/1.4 a lot