Formula - Bounce height with respect to time

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?

• Welcome to Physics! Please note that homework-like questions and check-my-work questions are generally considered off-topic. – ACuriousMind Nov 19 '15 at 15:17
• 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? – user83548 Nov 19 '15 at 15:38
• For one thing, you didn't actually answer the question. – Jeff Nov 19 '15 at 15:45
• @Jeff: please don't use emotive straw men. Bruce never said anything of the sort. – Gert Nov 26 '15 at 3:52
• He did, just deleted his comment. – Jeff Nov 27 '15 at 21:37

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.

• So, my first formula will be different from the subsequent bounces, right? and $v_{bounce}$ is the initial velocity leaving the ground, right? – Jeff Nov 19 '15 at 16:13
• just a clarification, H is the initial release height, not the height after each bounce. – user83548 Nov 19 '15 at 16:24
• Sorry, there was a typo in th previous expression, $v_{bounce}=e^n\sqrt{2H/g}$, for the $n^{th}$ bounce – user83548 Nov 19 '15 at 16:33
• 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) – user83548 Nov 19 '15 at 16:42
• You are welcomed, and I apologize if I sounded mean, it was not my intention. – user83548 Nov 20 '15 at 16:45