juggler keeps on moving four balls in the

air continuosly such that ball attains 20 m

height .when the first ball leaves the hand

the position of other balls in metre

Dear Sourav

 

Given are the datas,u = 20m/sv = 0 final velocitya = -g = -10m/s^2.  So, the time taken by ball to reach its maximum height 0 = 20-10tt = 2 sec.  So, total time will be 4 sec i.e. juggler will touch the same ball after 4 sec only. Since, he is juggling 4 balls at regular intervals, i.e. each of them is spaced at 1 sec distance interval. Also, the maximum height to which the ball reaches,From Newton’s third law of eqn. we get,v^2 = u^2 + 2ass = 20m Hope it helps  RegardsArun (askIITians forum expert)

T=2u/g

2.10/20=4

Therefore time interval is 1sec

H=20-1/2.10=15m

H'=20*3-1/2.10*9=15m

H''=20*3-1/2*10*4=20m