Question

# A man throws balls with the same speed vertically upwards one after the other at an interval of What should be the speed of the throw so that more than two balls are in the sky at any time?

Solution

Given:

The man throws a ball with the speed which equal for the entire ball and every ball is thrown at an interval of 2 seconds. ( T )

Solution:

The speed of the ball is calculated using first equation of motion, ( v=u+ )

at.

So, the equation formed in this situation will be more than 2 balls be in air means let us assume 3 balls so the time taken by first ball in flight be more than or equal to 4 seconds, as per the situation.

Therefore, we can say that when ( t=4 ) seconds the third ball is thrown.

Therefore the first ball equation be, ( v=u+a t )

Final velocity ( =-v, ) Initial velocity ( =-u, ) Acceleration due to gravity ( =-g ) ( -u=u-g t )

( u=frac{g t}{2} )

( u=9.8 times frac{4}{2} )

( u=19.6 mathrm{m} / mathrm{s} )