Calculating the height objects fall from 
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*Imagine an object of a given mass.

*The object falls from a certain height.

*On contact with the ground, the object has a certain force.
If the force of the object and the mass of the object are know, what formula would be used to find the height fallen from?
 A: Knowing only the force is not sufficient to calculate the original height. This is because in order to find this, one needs to calculate the velocity or momentum of the object at the instant before it collides with the ground. Consider how the motion of the object is affected as it collides with the ground. As soon as the object makes contact, the force of the ground acting upward on the object causes a deceleration, which presumably occurs until the object reaches a velocity of $0$. This situation is clearly dependent not only on the force applied but also the time that it was applied. Presumably, then, if you have not only the force, but the time over which the force was applied you could find the velocity with which the object struck the ground.
The way this works is by using the impulse, which is given by:
$$\int_{t_i}^{t_f} F\cdot dt $$This expression is equivalent to the change in momentum of the object:
$$\int_{t_i}^{t_f} F\cdot dt = p_f - p_i$$Because the final momentum is $0$ we have: 
$$\int_{t_i}^{t_f} F\cdot dt = - p_i$$Solving the integral gives us the momentum just before the object hit the ground. Using this, one could find the corresponding velocity by using the fact that $p=mv$. Finally, all it would take is a kinematic equation to find the initial height:
$$(v_f)^2-(v_i)^2 = 2a\Delta y $$Where $\Delta y$ is the desired height, $v_f$ is the quantity we previously found, and $v_i=0{m \over s}$ if the object begins falling from rest. Its worth noting that if the force of the ground on the ball is given as an average force, meaning it is the constant equivalent to the real force versus time expression, the impulse integral simplifies simply to $\Delta p =F_{avg}* \Delta t$.
