# A ball is thrown, where does it bounce?

Firstly sorry if this has been asked before. I'm trying to work out a simple graphical equation that'll give me the x of the bounce point when y=0 (as it's hit the ground).

I'm only after the first bounce and the peak of the second bounce (so essentially it's like throwing a ball at a wall, getting it's first bounce and the point at which it hits the wall).

Gravity and the weight of the ball will always be constant, there will be no air resistance and the only variables will be the height from which the ball is thrown and the speed it's thrown at.

Can you help me out?

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@Ahmed I do not fully understand your question. I am making a number of assumptions here; a better answer can be provided if you make the question clearer. I am assuming that you are throwing a ball horizontally with a velocity $u$ from a height $h$. (Acceleration to gravity is $g \approx 9.81$.) In this case, the $x$ coordinate of the first bounce is given by the formula $u \sqrt{\frac{2h}{g}}$. I am also assuming that the collision with the ground is perfectly elastic (i.e., it rebounds with the same speed). In this case, the ball will reach a maximum height of $h$ after the first bounce – Srivatsan Jul 18 '11 at 22:35
Let me make sure I understand. We will throw a ball with an initial height and velocity velocity (note that I mean both speed and direction). We assume no obstructions other than a perfectly flat ground, and any and all collisions will be perfectly elastic (the initial energy of the ball is conserved). Is that correct? – mixedmath Jul 18 '11 at 22:38
@Srivatsan: Why don't you try to answer or ask one or two questions to get the required 50 rep so that you can comment properly and not bother the moderators with the "comments posted as answers"? – t.b. Jul 18 '11 at 22:39
(this is Srivatsan's comment continued): is given by the formula $u \sqrt{\frac{2h}{g}}$. I am also assuming that the collision with the ground is perfectly elastic (i.e., it rebounds with the same speed). In this case, the ball will reach a maximum height of $h$ after the first bounce. – Zev Chonoles Jul 18 '11 at 23:15
Sorry for not explaining it correctly. I should have mentioned that I want some sort of gravity constant in there so that when it rebounds, it rebounds at a degraded height. The path of the ball will be on a 2D axis, so imagine the path across your screen going from left (where we throw it from), bouncing about 3/4 of the way and touching the right side of your screen. – Ahmed Nuaman Jul 19 '11 at 6:30

Suppose a ball is launched at time $t=0$ starting at $x=0$ and $y=y_0$ with initial velocisty $v_{0,x}$ and $v_{0,y}$. Then, after a time $t$, the $y$-coordinate of the ball will be (assuming constant gravitational force) $$y=-1/2gt^2+v_{0,y}t+y_0,$$ where $g$ is the (magnitude) of the acceleration due to gravity at the surface of the earth. We are interested when the ball lands on the ground, that is, when $y=0$. Setting $y=0$ into the above equation and solving for $t$, we find $$t=\frac{1}{g}\left( v_{0,y}+\sqrt{v_{0,y}^2+2gy_0}\right)$$ is the only positive root of the resulting equation. On the other hand, after a time $t$, the $x$-cooridnate of the ball will be $$x=v_{0,x}t.$$ Thus, to find the $x$-coordinate when the ball hits the ground, we need to merely plug in our result from above. We obtain $$x=\frac{v_{0,x}}{g}\left( v_{0,y}+\sqrt{v_{0,y}^2+2gy_0}\right) .$$

At the peak of the bounce, the $y$-coordinate will not be changing, that is $y'(t)=0$, which give us the following equation: $$0=-gt+v_{0,y}.$$ Solving for $t$ yields: $$t=v_{0,y}/g.$$ To find the height of the ball at this time we just plug this $t$ value into $y(t)$: $$y=\frac{-v_{0,y}^2}{2g}+\frac{v_{0,y}^2}{g}=\frac{v_{0,y}^2}{2g}.$$ Of course, under the assumption that the ball loses no energy after the first bounce, this will also be the height of the ball at the second peak.

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Amazing! Will pipe this into my code tonight :) – Ahmed Nuaman Jul 19 '11 at 6:33

Instead of telling you the answer directly, I will comment on how you should go about solving this question.

The initial idea is to find the vertical velocity of the ball. Once you isolate this dimension, it's a 1D kinematics problem. The only formulas that you will need are

\begin{align} s &= \frac{1}{2}a\cdot t^2 + v_0\cdot t + s_0 \\ v &= a\cdot t + v_0 \\ a &= a \end{align}

Here's a little trick that I like to do. We can quickly find out how high the ball reaches using energy (take the initial vertical speed and height and calculate the starting energy: $1/2 \cdot mv^2 + mgh$. Solve for height H by setting equal to $mgH$. Now we have the height. How long does it take to hit the ground? One could solve for $\frac{1}{2} m V^2$ to see bottom velocity and then use $v = aT$ to do it entirely in your head. No energy is lost, so the time it takes for one parabolic path is $2T$. This quickly allows us to solve for any number of bounces, and with only mental computation.

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Looks like I got me some homework! Thanks, I'll try my best :) – Ahmed Nuaman Jul 19 '11 at 6:31
@Ahmed: no problem. Good luck! – mixedmath Jul 19 '11 at 6:53