A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws $n$ balls per second, the maximum height the balls can reach is

As the juggler is throwing $n$ balls each second and $2^{\mathrm{nd}}$ when the first is at its highest point, so the time taken by one ball to reach the highest point, $t=\frac{1}{n} \mathrm{s}$.

And as at highest point $v=0$.

So, from first equation of motion,

$$\begin{gathered} v=u+at \\ \Rightarrow 0=u-g\left(\frac{1}{n}\right) \\ \Rightarrow u=\frac{g}{n} \end{gathered}$$

Now in order to calculate height, using second equation of motion, we have:
  $\text{S}=ut+\frac{1}{2}{\text{at}}^2$      ...(i)

$\Rightarrow h=\frac{1}{n}\times \frac{g}{n}-\frac{1}{2}\text{g}\times {\left(\frac{1}{n}\right)}^2$

     $\Rightarrow h=\frac{g}{2n^2}$  

So, the balls rise up to a height of $h=\frac{g}{2n^2} \mathrm{m}$.