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

(c) We can write `(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)`
`2 cot theta=sqrt(3)impliestan theta=2/(sqrt(3)), theta=tan^(-1)(2/(sqrt(3)))`