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)((4R)/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 cos theta)/g.........(ii)`
Dividing (i) by (ii), we get
`H/R=(tan theta)/4 or tan theta=(4H)/R or theta=tan^(-1) ((4H)/R)`