Calculating Average Force in a Ball Rebound Momentum Problem

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The discussion focuses on calculating the average force exerted by the ground on a 0.3 kg ball dropped from a height of 2 meters, which rebounds to a height of 1.5 meters. The collision duration is 0.01 seconds. The correct average force is determined to be 350 N, calculated using the formula involving gravitational potential energy (mgh) and the change in velocity during the impact. The equation used was (0.3 kg * (√(2 * 9.81 m/s² * 2 m) + √(2 * 9.81 m/s² * 1.5 m))) / 0.01 s.

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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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