A juggler keeps on moving 4 balls in air in regular intervals of time.when one ball leaves his hand(speed=20m/s),position of other three balls will be?

So, the initial speed of the ball as it leaves the jugglers hand is 20m/s 

u = 20m/s

v = 0 (Since the ball will reach 0 velocity, once it reaches the maximum height possible and then will fall back again)

a = -g = -10m/s^2. 

 

So, the time taken by ball to reach its maximum height 

v = u+at

0 = 20-10t

t = 2 sec. 

 

So, total time for which the ball will be in air 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

v^2 = u^2 + 2as

400 = 20s

s = 20m 

 

hence, currently, when one ball leaves juggler's hand, other balls will be at 10 (going up), 20 and 10 (coming down) m respectively.