From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is

Options

A. `gH=(n-2)u^(2)`
B. `2gH=n^(2)u^(2)`
C. `gH,=(n-2)^(2)u^(2)`
D. `2gH=nu^(2)(n-2)`

Answer

D

Explanation

(d): Time taken to reach the maximum height h is `t_(1)`
`v=u+"at"_(1)`
`0=+u-"gt"_(1)`
`t_(1)=(u)/(g)`

Let time taken to hit the ground be `t_(2)`
`-H=+"ut"_(2)-(1)/(2)g t_(2)^(2)`
Given , `t_(2)="nt"_(1)`
`-H="unt"_(1)-("gn"^(2))/(2) t_(1)^(2)`
`="un"(u)/(g)-("gn"^(2))/(2)(u^(2))/(g^(2))`
`-H=(u^(2)n)/(g)-(n^(2)u^(2))/(2g)`
`=(nu^(3))/(g)(1-(n)/(2))`
`2g H=nu^(2) (n-2)`