Figure shows the trajectory of a projectile fired at an angle `theta` with the horizontal. The elevation angle of the highest point as seen from the point of launching is `varphi`. The relation between `varphi` and `theta` is :

Options

A. `tanvarphi=1/2tantheta`
B. `tan^(2)varphi=1/2tan^(2)theta`
C. `sinvarphi=1/2sintheta`
D. `cos^(2)varphi=1/2cos^(2)theta`

Answer

A

Explanation

Here `tanphi=H/(R/2)=((u^(2)sin^(2)theta)/(2g))/((2u^(@)sinthetacostheta)/(2g))=(sintheta)/(2costheta)`
`tanphi=1/2tantheta`