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 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time (Given g = $10m/s^2$)

• 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.

1st ball caught 2nd 3rd 4th in air

So time of flight of 1^st ball is 6 sec. it took 6s for the first ball to be caught.

This is the limiting case. T$\ge$ 6 sec

Let the velocity with which it was thrown be u

Acceleration = - $10m/s^2$

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

Since ball comes back to starting position

So, s = 0

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

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

We get u$\ge$ 30 m/s