A juggler keeps on moving four balls in the air throws the balls in regular interval of time. When one ball leaves his hand (speed = 20 m/s), the positions of other balls will be (take g = 10 m/s²)

The time taken by the ball to reach its maximum height is given by
v=u-gt or 0 =20 -10 x t
$\therefore$ t=2s
The positions of different balls are shown in figure

The maximum distance travelled in given by 
$\mathrm{h}=\mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^2$
or $\mathrm{h}=20\times 2-\frac{1}{2}\times 10\times (2{)}^2=20\mathrm{m}$
Distance travelled by first ball
$\begin{array}{l}{\mathrm{h}}_1=\frac{1}{2}{\mathrm{gt}}^2=\frac{1}{2}\times 10\times (1{)}^2=5\mathrm{m}\\ {\mathrm{h}}_2=\frac{1}{2}{\mathrm{gt}}^2=\frac{1}{2}\times 10\times (2{)}^2=20\mathrm{m}\end{array}$
The distances from ground a.re
15m,20m,15m