A juggler throws balls vertically upwards with same initial...

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

And as at highest point $v = 0$.

So, from first equation of motion,

$v = u + at ⇒ 0 = u - g\left(\dfrac{1}{n}\right) ⇒ u = \dfrac{g}{n}$

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

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

$⇒ h = \dfrac{g}{2{n}^{2}}$

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