Question

A person throws vertically $10$ 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. $10\text{m}$
• B. $5\text{m}$
• C. $5\text{cm}$
• D. $10\text{cm}$

Answer 1

The correct option is C $5\text{cm}$

$10$ balls are thrown per second.
Time interval between two balls thrown $=\frac{1}{10}\text{s}$
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
$\Rightarrow 0=u-g\left(\frac{1}{10}\right)$ (By first equation of motion)
$\Rightarrow u=\frac{g}{10}$
Height achieved by the balls is given by
$v^2-u^2=2aS$
$\Rightarrow 0^2-{\left(\frac{g}{10}\right)}^2=2\times (-g)h$
$\Rightarrow h=\frac{g}{2\times {10}^2}=0.05\text{m}=5\text{cm}$
The answer is (c)