If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is

Options

A. `tan^(-1)((H)/(R ))`
B. `tan^(-1)((2H)/(R ))`
C. `tan^(-1) ((4H)/(R ))`
D. `tan^(-1) ((4H)/(H))`

Answer

C

Explanation

Maximum height, `H = (u^(2) sin^(2) theta)/(2g) " "…(i)`
Horizontal range, `R = (u^(2) sin 2theta)/(g) = (2u^(2)sin theta costheta)/(g) " "...(ii)`
Divide (i) by (ii), we get
`(H)/(R ) = (tan theta)/(4) or tan theta = (4H)/(R ) or theta = tan^(-1) ((4H)/(R ))`