Calculating Average Force in a Ball Rebound Momentum Problem

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To calculate the average force exerted by the ground on a 0.3 kg ball dropped from 2 m and rebounding to 1.5 m, the momentum change during the collision must be considered. The potential energy at the drop height (mgh) converts to kinetic energy before impact, and the rebound height indicates the energy retained. The average force can be calculated using the formula involving the change in momentum divided by the collision time of 0.01 seconds. The calculations confirm that the average force is 350 N, which matches one of the provided options. This approach effectively demonstrates the relationship between energy, momentum, and force in rebound scenarios.
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A .3 kg ball is dropped from a height of 2 m. It strikes the group and rebounds to a height of 1.5 m. The collision with the ground lasts for .01 seconds. The average force exerted by the ground on the ball is:

a)280
b) 540
c) 430
d) 350

So, i know that we have to use some type of formula involving mgh which equals velocity?

I did:

.3 [ rad 2mgh +rad 2 mgh] / .01 = 350

is this correct?
 
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