A jugglar maintains four balls in motion, vaking each in turn rise to a height of `20 .0 m` from his hand .Find the velocity with which the jugglar project these balls and the position of other three balls at the instant when the foruth ball is just leaving the hand of jugglar. Take `g= 10 m//s^2.

Explanation

Taking vertical upward motion of a ball, we have
` S= 20 m , a=- 10 m//s^2 , ltbRgt ` u?, v=0, t=?`
As ` v^2 =u^2 + 2 aS` so, ` 0=u^2 =2 ( -10) xx 20`
or ` u = 20 ms^(-1)` ltbRgt Also ` v=u + at ` or ` t=(v-u) /a = )0-20)/(-10) =2s`
so each ball will return to the hand of jugglar after ` 4 s`.
To maintain proper distance, the balls must be projected upwards with an interval of time
` = 4/4 = 1 s`. When the fourth ball is in hand, the third ball has travelled for ` 1 s`, second ball fro ` 2` s and dirst ball for 3 s`.
(i) Fro Third ball, ltbRgt ` S= ut + 1/2 at^2 = 20 xx 1 + 1/2 (-10) 1^2 = 15 m`
Thus third ball will be ` 15 m` ablove the ground going upwards.
(ii) For second ball,
` S =20 xx 2 + 1/2 ( -10) xx 2^2 = 20 m`
The second ball will be ` 20 m` abover the ground just at rest.
(iii) For first ball,
` S= 20 xx 3 ss 1/2 + ( -1 0) xx 3^2 = 15 m`
The first ball will be ` 15 m above the ground moving dounwards.