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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?

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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 :)

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  • $\begingroup$ Readers should note that $\Delta t$ in this case is the impact timescale, which is a small number (e.g., less than 1 sec). $\endgroup$ – Kyle Kanos Dec 9 '17 at 12:49
  • $\begingroup$ Yes, but why is that so? It seems that the force should be $ma$. $\endgroup$ – Sillysack Buttowski Dec 9 '17 at 12:51
  • $\begingroup$ @Sillysack This is still $F=ma$. It is the force of the impact so the acceleration in the equation pertains to the change in speed of the impact. $\endgroup$ – CooperCape Dec 9 '17 at 12:55
  • $\begingroup$ Alright, how do we find out the time taken for the impact then? I don't think anyone with a stopwatch could time that... how do we calculate the time then? $\endgroup$ – Sillysack Buttowski Dec 9 '17 at 13:00
  • $\begingroup$ @Sillysack I suppose a simple way is to film it and look at the video frame by frame and get a pretty good estimate from that. $\endgroup$ – CooperCape Dec 9 '17 at 13:09
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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.

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  • $\begingroup$ "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". How can the object have an upward acceleration when it is, in fact, moving downward? And how does it vary with time? $\endgroup$ – Sillysack Buttowski Dec 9 '17 at 14:23
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    $\begingroup$ @SillysackButtowski Have you studied any physics? Acceleration is the change in velocity. It doesn't tell us what direction the object is moving. $\endgroup$ – Bill N Dec 9 '17 at 14:57
  • $\begingroup$ @NathanaelVetters I don't see anything wrong with what you said. There is a large upward acceleration while the object is rebounding off the floor. $\endgroup$ – Bill N Dec 9 '17 at 15:04
  • $\begingroup$ If I am moving and slowing down, I am in fact accelerating in the opposite direction of my velocity. Acceleration can be in any direction regardless of velocity. $\endgroup$ – Nathanael Vetters Dec 9 '17 at 15:05
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    $\begingroup$ @SillysackButtowski Here's a website you should gradually work your way through. It's full of information at a foundational level. hyperphysics.phy-astr.gsu.edu/hbase/hframe.html $\endgroup$ – Bill N Dec 9 '17 at 15:05

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