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

Options

A. `(g)/(2(n -1)^(2))`
B. `(g)/(2n^(2))`
C. `(g)/(n^(2))`
D. `(g)/(n)`

Answer

B

Explanation

Time interval between two balls = Time of ascend
`= (1)/(h) = (u)/(g) u = (g)/(h) , h = (u^(2))/(2g)`