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A juggler sets 4 balls in motion in succession to a height of 19.6 m from his hand. What is the velocity of projection of each ball? Once set in motion, what are the displacements of the other three balls when the 2nd ball just leaves his hand?

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Solution

Hmax=19.6m
v2u2=2as
At maximum height, v=0
u2=2as
u=2gHmax
u=2×9.8×19.6
u=19.6m/su=19.6m/s^j
Time of flight of each ball =2ug=2×19.69.8=4s
We can see,juggler has to manage 4 balls each having time of flight 4s
He has to manage the time in this symmetrically way to get his job done.
S1 =displacement of first ball
S2 =displacement of third ball
S1 =displacement of fourth ball
S1=ut12gt2
=19.6×112×9.8×1=14.7m
S1=14.7^j(m)
S3=19.6×212×9.8×4
S3=19.6^j(m)
S4=19.6×312×9.8×9
S4=14.7^j(m)

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