Question

A boy throws $n$ balls per second at regular time intervals. When the first ball reaches the maximum height he throws the second one vertically up. The maximum height reached by each ball is

• A. $\frac{g}{2(n-1{)}^2}$
• B. $\frac{g}{2n^2}$
• C. $\frac{g}{n^2}$
• D. $\frac{g}{n}$

Answer 1

The correct option is B $\frac{g}{2n^2}$

Time gap between two successive throws is $t=\frac{1}{n}$

so it is the time taken by one ball to reach the top point.

$\text{In other words it is the time taken to make the velocity zero so we can calculate the velocity of projection.}$

If velocity of projection is $u$ then final velocity at top point(comes to rest) i.e $v=0=u-gt=u-g\left(\frac{1}{n}\right)$

so $u=g/n$

Now we know that for projection velocity, $u$ the maximum height is $h=\frac{u^2}{2g}=\frac{g^2}{n^2\times 2g}=\frac{g}{2n^2}$

Option B is correct.