A projectile is thrown at an angle `theta` that it is just able to cross a vertical wall at its highest point of journey as shown in the figure. The angle `theta` at which the projectile is thrown is given by

Options

A. `tan^(-1) ((1)/(sqrt(3)))`
B. `tan^(-1)sqrt(3)`
C. `tan^(-1)((2)/(sqrt(3)))`
D. `tan^(-1)((sqrt(3))/(2))`

Answer

C

Explanation

From the given diagram,
`(R//2)/(H) = (sqrt(3)H)/(H) = sqrt(3)`
or `((v_(0)^(2)sin theta cos theta)//g)/((v_(0)^(2)sin^(2)theta)//2g) = sqrt(3) rArr 2 cot theta = sqrt(3)`
or `tan theta = (2)/(sqrt(3))` or `theta = tan^(-1)((2)/(sqrt(3)))`