A juggler keeps on moving four balls in air throwing the balls after regular intervals. When one ball leaves his hand (speed = 20 m s^-1), the position of other balls (height in
meter) will be (take g = 10 ms^-2)

Time taken by the same ball to return to the hands of the juggler is $\frac{2u}{g}=\frac{2\times 20}{10}=4\mathrm{s}$. 

So he is throwing the balls after 1s each. Let at some instant he throws ball number 4. 

Before 1 s  of throwing it, he throws ball 3. So the height of ball 3 is 

$h_3=20\times 1-\frac{1}{2}10(1{)}^2=15\mathrm{m}$

Before 2 s, he throws ball 2. So the height of ball 2 is

$h_2=20\times 2-\frac{1}{2}10(2{)}^2=20\mathrm{m}$

Before 3 s, he throws ball 1. So the height of ball 1 is

$h_1=20\times 3-\frac{1}{2}10(3{)}^2=15\mathrm{m}$