Calculating Distance Using Force, Mass, and Acceleration

bradosaurus

1. The problem statement, all variables and given/known data


A rock is dropped from x(meters) height. The rock weighs 68 kg, and its force at impact with the ground is 15 kilonewtons. If the acceleration due to gravity is 9.8m/s^2, what is the distance/height the rock has fallen?


2. Relevant equations


F=m*a

15kN=15000kg*m/s^2

666.4N=68kg*9.8m/s^2

The force exerted by the rock at stationary position is 666.4 Newtons.

The rock fall distance is x meters.


3. The attempt at a solution


666.4N*x=15000N

(666.4kg*m/s^2)*x=15000kg*m/s^2

x=(15000kg*m/s^2)/(666.4kg*m/s^2)

x=22.51


This was all I could come with, but it is simply the ratio of the force. How do I find distance?

Doc Al

Please state the complete problem exactly as given. As you've written it, there's not enough information. (The impact force depends on the nature of the surface being hit.)

AC130Nav

Shouldn't there be a formula for the kinetic energy of a mass falling distance x and a formula for the F generated by an unknown kinetic energy when it hits? The two could then be equated resolving x.

I don't see elastic/inelastic collisions mentioned.

Doc Al

That one's easy.
That one's not so easy. More information is needed.

bradosaurus

Thank you for your responses. This was a question posed to me by a friend of mine who rock climbs. He inferred that the distance of the fall could be calculated using the force, mass and gravitational constant. Is it correct to say that the energy spent to stop will be the same for different stopping distances while the required force will be variable?

The question may be better understood as follows:

A person who weighs 68 kg falls an unknown distance. The climber's rope has a breaking strength of 15 kN, and a dynamic elongation of 29%. Barring any other variables (friction, etc.), what is the furthest distance the climber could fall before the rope fails?