Question

A man throws balls (and catches at the same height) with the same speed vertically upwards one after the other at an interval of $2\text{s}$. What should be the speed of the throw so that more than two balls are in the sky at any time (Given $g=10{\text{m/s}}^2$)

• A. At least 20 m/s
• B. Any speed less than 30 m/s
• C. Only with speed 30 m/s
• D. More than 30 m/s

Answer 1

The correct option is D

More than 30 m/s

3 balls have to be in air at all times.

So when the 1st ball is caught, 2nd 3rd 4th are still in air.

Now, the time of flight of the 1^st ball is $6\text{s}$. it took $6\text{s}$ for the first ball to be caught after it was released. (The man throws the ball at an interval of $2\text{s}$)

This is the limiting case. T$\ge 6\text{s}$

Let the velocity with which it was thrown be $u$

Acceleration = $-10{\text{m/s}}^2$

$s=ut+\frac{1}{2}a{t}^2$

Since ball comes back to starting position

So, $s=0$

$ut-\frac{1}{2}g{t}^2=0$

$u=\frac{1}{2}gt$
Since $t\ge 6$

We get $u\ge 30\text{m/s}$