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 byv=u-gt or 0 =20 -10 x t∴ t=2sThe positions of different balls are shown in figureThe maximum distance travelled in given by h=ut−12gt2or h=20×2−12×10×(2)2=20mDistance travelled by first ballh1=12gt2=12×10×(1)2=5mh2=12gt2=12×10×(2)2=20mThe distances from ground a.re15m,20m,15m