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 inmeter) will be (take g = 10 ms-2)

Time taken by the same ball to return to the hands of the juggler is 2ug=2×2010=4s. 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 h3=20×1−1210(1)2=15mBefore 2 s, he throws ball 2. So the height of ball 2 ish2=20×2−1210(2)2=20mBefore 3 s, he throws ball 1. So the height of ball 1 ish1=20×3−1210(3)2=15m