A boy throws n balls per second at regular time intervals. When the first ball reaches the maximum height he throws …..

Q. A boy throws n balls per second at regular time intervals. When the first ball reaches the maximum height he throws the second one vertically up. The maximum height reached by each ball is

(a) \displaystyle \frac{g}{2(n-1)^2}

(b) \displaystyle \frac{g}{2n^2}

(c) \displaystyle \frac{g}{n^2}

(d) \displaystyle \frac{g}{n}

Ans: (b)
Sol:
Time taken by ball to go maximum height
t = 1/n

v = u – gt

0 = u – g(1/n)

u = g/n = velocity with which ball was projected

Maximum height reached by ball is

\displaystyle H = \frac{u^2}{2g}

\displaystyle  H = \frac{g}{2n^2}

Author: Rajesh Jha

QuantumStudy.com

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