1
$\begingroup$

I'm trying to graph a bounce with respect to time. I have these formulas:

$\frac 12 mV_2^2 = mgH_2$

and

$H_2 = \frac{1}{2} \frac{V_2^2}{g}$

I will have a series of $H(t)$ formulas as I know how to get the next bounce's initial velocity (last bounce's final velocity * -(coefficient of restitution)).

Is there a way to get an H(t) formula from these? I am not a physicist, but a computer science teacher. I have thoughts, but I am not sciencey enough to figure it out.

Thoughts:

Velocity is distance / time (position? / time). Would that mean that:

$\frac 12 mV_2^2 = mgH_2$ =>

$mV_2^2 = 2mgH_2$ =>

$V_2^2 = 2gH_2$ =>

$V_2 = \sqrt{2gH_2}$ =>

$d/t = \sqrt{2gH_2}$ =>

$d = \frac{\sqrt{2gH_2}}{t}$

Does that work?

Improve this question
$\endgroup$
5
  • $\begingroup$ Welcome to Physics! Please note that homework-like questions and check-my-work questions are generally considered off-topic. $\endgroup$
    – ACuriousMind ♦
    Commented Nov 19, 2015 at 15:17
  • $\begingroup$ it doesnt work because the speed is not constant, so you cannot define it as $d/t$. I already pointed you to the correct equation in your previous question and gave you a reference, what part did you do not understand? $\endgroup$
    – user83548
    Commented Nov 19, 2015 at 15:38
  • 1
    $\begingroup$ For one thing, you didn't actually answer the question. $\endgroup$
    – Jeff
    Commented Nov 19, 2015 at 15:45
  • $\begingroup$ @Jeff: please don't use emotive straw men. Bruce never said anything of the sort. $\endgroup$
    – Gert
    Commented Nov 26, 2015 at 3:52
  • $\begingroup$ He did, just deleted his comment. $\endgroup$
    – Jeff
    Commented Nov 27, 2015 at 21:37

1 Answer 1

Reset to default
2
$\begingroup$

you have to notice that this motion is accelerated, so if you define velocity as $d/t$ you will get the wrong result. For uniformly accelerated motion (which is your case, the acceleration is contant: g), you have to use the following relationship:

$y(t)=y_0+v_0t+\frac{1}{2}at^2$

When you release the ball, $y_0=H$, $v_0=0$ and $a=-g$ so you get $ y(t)=H-\frac{1}{2}gt^2$. After the first bounce you get, for each bounce: $y(t)= v_{bounce}t-\frac{1}{2}gt^2$

The graphic is a series of inverted parabolas, there is no single formula to encompass the entire curve. You have to define it piecewise.

Improve this answer
$\endgroup$
5
  • $\begingroup$ So, my first formula will be different from the subsequent bounces, right? and $v_{bounce}$ is the initial velocity leaving the ground, right? $\endgroup$
    – Jeff
    Commented Nov 19, 2015 at 16:13
  • $\begingroup$ just a clarification, H is the initial release height, not the height after each bounce. $\endgroup$
    – user83548
    Commented Nov 19, 2015 at 16:24
  • $\begingroup$ Sorry, there was a typo in th previous expression, $v_{bounce}=e^n\sqrt{2H/g}$, for the $n^{th}$ bounce $\endgroup$
    – user83548
    Commented Nov 19, 2015 at 16:33
  • $\begingroup$ Th reason being that the ball returns to the ground with the same speed it left, and after the bounce will be reduced by a factor $e$ (coefficient of restitution, not the number e) $\endgroup$
    – user83548
    Commented Nov 19, 2015 at 16:42
  • $\begingroup$ You are welcomed, and I apologize if I sounded mean, it was not my intention. $\endgroup$
    – user83548
    Commented Nov 20, 2015 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.