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

Step-by-step solution

Time taken by the small ball to return to the hands of the juggler is 2 ~V g = 2 20 10 =4 ~s . So, he is throwing the balls after 1 s each. Let at some instant, he throws the ball number 4 . Before 1 s of throwing it, he throws ball 3 . So, the height of ball 3 is aligned h & =u t- 1 2 g t^2 h₃ & =20 1- 1 2 aligned Before 2 s , he throws ball 2 . So, the height of ball 2 is h₂=20 2- 1 2 10(2)^2=20 ~m Before 3 s , he throws ball 1 . So, the height of ball 1 is aligned & h₁=20 3- 1 2 10(3)^2 & h₁=15 ~m aligned