A boy throws balls into air.He throws one, whenver the previous one is at its highest point. How high do the balls rise if he throws one ball each sec ?

Options

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

Explanation

A juggler is throwing n balls each second and` 2^(nd)` when the first is at its highest point, so time taken by one ball to reach the highest point t = (1/n) sec and as at highest point v = 0,
From 1st equation of motion
0 = u – (g) (1/n), i.e. u = (g/n) .....(A)
Now from 3 rd equation of motion
i.e. ` v^(2) = u^(2) + 2as, 0 = u^(2) - 2gh`
i.e. ` h= (u^(2) //2g)`
` h = (g)/(2n^(2)) [ " From Eq. (1) " u= (g)/(n)]`
Hence correct answer is (B)