Question
A juggler throws balls into air. He throws one whenever the previous one is at its highest point. How high do the balls rise if the throws n balls each sec? Acceleration due to gravity is g.

None of these



easy
Solution
As the juggler is throwing n balls each second and 2^{nd} when the first is at its highest point, so the time taken by one ball to reach the highest point,
t = (1/n)s
and as at highest point v = 0, from 1^{st} equation of motion,
0 = u – (g)(1/n), i.e., u = (g/n) …(1)
Now from 3^{rd} equation of motion, i., v^{2} = u^{2} + 2as,
0 = u^{2} – 2gh, i.e., h = (u^{2}/2g)
SIMILAR QUESTIONS
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