Calculating Force of a Free Falling Weight: 500 lb Steel Ball

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SUMMARY

The discussion focuses on calculating the force produced by a 500 lb steel ball when dropped from a height of 3 ft, with inquiries about the force at greater heights of 6 ft and 9 ft. The formula for force in this context is derived from the impulse-momentum theorem, specifically considering perfectly elastic collisions. The force can be calculated by determining the change in momentum of the ball divided by the time duration of the collision. The force does not simply double or triple with height; it is determined by the square root of the height due to gravitational acceleration.

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Dale Murphy
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can someone help me with this.
you have a free falling weight such as a 500 lb. steel ball dropping from 3 ft. How much force will the weight produce in ft. lbs. or tons when hitting the ground. what is the formula for figuring this. Also, will the amount of force double if dropped from 6 ft? Triple from 9 ft.?
Thanks
 
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this comes under the topic of impulse and collisions.
http://en.wikipedia.org/wiki/Impulse_(physics )
http://en.wikipedia.org/wiki/Collision

here, there are several cases to be considered as in real life. But assuming the collision is perfectly elastic, then the impulse of the ball would be the change in the momentum just before and after the collision. This divided by the fractional time taken for the collision would give you the force.
 
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