A body is projected vertically upwards with a velocity u from the top of a tower. Time taken by it to reach the ground is n times, then the time taken by it to reach the highest point in its path. Height of the tower is

Options

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

Answer

C

Explanation

Let the time taken to reach the maximum height, when thrown vertically upwards
`t_(1) = (u)/(g)`
If `t_(2)` be the time to hit the ground, then given
`t_(2) = nt_(1) = nu//g`

So, using equation of motion for distance,
i.e., `s = ut + (1)/(2) "at"^(2)`
`-H = u.n ((u)/(g)) - (1)/(2) g(n^(2) u^(2))/(g^(2)) [a = -H, t = t_(2) = nt_(1)]`
or `-H = (nu^(2))/(g) - (n^(2) u^(2))/(2g)`
or `2gH = nu^(2) (n-2)`
or `H = (nu^(2) (n-2))/(2g)`