A projectile is fired from the base of coneshaped hill. The projectile grazes the vertex and strikes the hill again at the base.If `alpha` be the half-angle of the cone, `h` its height, `u` the initial velocity of projection and `theta` angle of projection, then then `theta tan theta`is

Answer

B

Explanation

Here range `=2h tan alpha=(u^(2)sin 2theta)/g` and `h=(u^(2)sin^(2) theta)/2g` Dividing `2tan alpha=(2sin 2theta)/sin^(2)theta rArr tan alpha=2cot theta`
`:.u^(2)(2gh)/(sin^(2)theta)=(2gh)/(((2cotalpha)/sqrt(1+4cot^(2)alpha)))=(2gh(1+4cot^(2)alpha))/(4 cot^(2)alpha)=gh(1/2tan^(2)alpha+2)`
`tan theta=2cot alpha`