${\displaystyle \stackrel{··}{y}=g}$

where $y$ is measured upwards. Integrating wrt $t$

${\displaystyle \stackrel{·}{y}=v=u+\mathrm{gt}}$

Integrating a second time and taking $y=0$ when $t=0$

${\displaystyle y=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2}$

Eliminating $t=\frac{v-u}{g}$ we get

${\displaystyle y=\frac{u(v-u)}{g}+\frac{g}{2}\frac{(v-u{)}^2}{g^2}}$

which simplifies to

${\displaystyle 2\mathrm{gy}=v^2-u^2.}$

This is equivalent to the energy equation for a mass $m$ where the change in potential energy in falling a distance $y$ is equal to the change in kinetic energy given by the equation:

${\displaystyle \mathrm{mgy}=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{mu}}^2}$

Now suppose the balls are dropped from rest from heights $H$ and $h$ then the lower one hits the ground after time

${\displaystyle t_1=\sqrt{\frac{2h}{g}}}$

and will bounce back to its initial position in an equal time. For the other ball to hit the lower ball before it falls for the second time the difference in initial heights must be such that the upper balls falls at least this far in time $2t_1$, that is

${\displaystyle H-h\ge \frac{g}{2}(2t_1{)}^2=\frac{g}{2}\times 4\frac{2h}{g}=4h.}$

Hence the 'top' ball must fall from no more than 4 times the height of the lower ball if they are to meet before the lower ball starts to fall a second time.