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

Answer

C

Explanation

Speed on reaching ground
`v= sqrt(u^(2) + 2gh)`
Now, `v = u + a t`
`rArr sqrt(u^(2) + 2gh) = -u + g t`

Time taken to reach highest point is `t= (u)/(g)`,
`rArr t= (u + sqrt(u^(2) + 2gH))/(g) = (n u)/(g)` (from question)
`rArr 2gH = n (n-2) u^(2)`