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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?

Asked by Karen Immanuel 28th March 2013, 8:04 PM
Answered by Expert
Answer:
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. 
Answered by Expert 12th May 2013, 4:45 AM
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