Question

juggler keeps on moving 4 balls in the air continously such that each ball attains 20m height. when the first ball leaves his hand the ppsition of the other ball will be

Answer 1

Given are the datas,u = 20m/s v = 0 final velocity a = -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

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