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

We have a juggler who throws $$n$$ balls per second, each with the same initial velocity $$u$$. He throws the next ball when the first ball reaches its highest position. The time interval between successive throws is $$\frac{1}{n}$$ seconds.

At the highest point the ball's velocity is zero, and the time to reach the maximum height is given by $$v = u - gt$$, so $$0 = u - g \cdot \frac{1}{n}$$, which gives $$u = \frac{g}{n}$$.

Now the maximum height is $$H = \frac{u^2}{2g} = \frac{1}{2g}\left(\frac{g}{n}\right)^2 = \frac{g^2}{2g n^2} = \frac{g}{2n^2}$$.

Hence, the correct answer is Option 4.