A ball is thrown, where does it bounce?

Question

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?

Answer 1

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.

Answer 2

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.