Force on a ball hitting the ground with acceleration $g$ at varying heights If a ball of mass $m$ falls from a height $h_1$ from rest and hits the ground after $t_1$ seconds at a velocity of $\sqrt{2gh_1}$, then the force acting on the ball will be $F=ma$ right?
Assuming that is correct, if the ball were to fall from a height $h_2$ where $h_2>h_1$, how would it be possible for the force to remain $F=ma$? Clearly, a ball thrown from a greater height would experience a greater force on impact right? What is going on here?
 A: The accleration in the equation $F=ma$ is not the acceleration due to gravity. It is easier to think about this by turning $F=ma$ into $$F=\frac{\Delta p}{\Delta t}$$Here we see that the acceleration in the equation is in fact the change in momentum of the object as it hits the ground. This can be re-written as $$F=\frac{mv-mu}{\Delta t}$$Where $v$ is the final speed of the object (in this case $0$), and $u$ is the initial speed of the object (in this case, the speed just before impact, or, as you have written $\sqrt{2gh}$).
Hope this helps in your understanding :)
A: During impact with the ground, "a" is no longer the acceleration due to gravity. It is the net acceleration that the object is currently experiencing. At the point of impact it now has a massive upward acceleration which is difficult to measure and varies during the course of the impact (which of course is quite a short time). If dropped from a greater height, this upward acceleration will be higher, and it might remain in contact with the ground for slightly longer. So $F = ma$ holds true, though it's not very useful for calculations. In this type of scenario, any calculations you do are much more likely to deal with momentum. But that is irrelevant to the question so I won't go into it in detail unless you so desire.
