A juggler keeps four balls in air. He throws each ball vertically upwards with the same speed at equal interval of time. The maximum height attained by each ball is $20 m$. Find kinetic energy of first ball when fourth ball is in hand. Assume that the mass of each ball is $10 \mathrm{grams}$

Let the time of descent of each ball is $t_0$

$\therefore h=\frac{1}{2}g{t}_0^{2}or\frac{20\times 2}{10}=t_0^2\therefore t_0=2s$

$\therefore$ Each ball remains in air for four seconds.  

Thus, the time interval between balls is 1 s

Therefore when 4th ball is in the hand, first ball is in descent and 1 second passed since its descent started.

The speed of first ball is  $v=gt=10\times 1=10m/s$

$\therefore$ The kinetic energy of the first ball is  

$KE=\frac{1}{2}m{v}_0^2=\frac{1}{2}\times 10\times {10}^{-3}\times 100=0.5J$