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

Answer

c

Explanation

Given that the height of the tower is H and the particle is thrown vertically upwards with a speed u.
Let the time taken by the particle to reach the highest point from the tower be t.
At the highest point the velocity of the particle becomes zero.
`0=-u+gtrArr t=(u)/(g)`
Then the time taken by the particle to hit the ground will be, `t_(1)` =nt.
From equation i
`t_(1)=(nu)/(g)`
`H=-ut_(1)+(1)/(2)gt_(1)^(2)`
From equation ii
`H=-unxx(u)/(g)+(1)/(2)gn^(2)((u)/(g))^(2)`
`H=-(u^(2)n)/(g)+(n^(2)u^(2))/(2g)`
`2gH=-2u^(2)n+n^(2)u^(2)`
`2gH=u^(2)n(n-2)`