Question

A person throws vertically $n$ balls per second with the same velocity. He throws a ball whenever the previous one is at its highest point. The height to which the balls rise is

• A. $\frac{g}{n^2}$
• B. $2gn$
• C. $\frac{g}{2n^2}$
• D. $2g{n}^2$

Answer 1

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

$n$ balls are thrown per second.
Time interval between two balls thrown $=\frac{1}{n}\text{sec}$
This is the time that a ball takes to reach its highest point
Using first equation of motion,
$v=u+at$
As the velocity of ball is $0$ at its maximum height
$\Rightarrow 0=u-g\left(\frac{1}{n}\right)$
$\Rightarrow u=\frac{g}{n}$
Height achieved by the balls is given by
$v^2-u^2=2as$
$\Rightarrow 0^2-{\left(\frac{g}{n}\right)}^2=2\times (-g)h$
$\Rightarrow h=\frac{g}{2n^2}$