For a projectile thrown with a velocity v, the horizontal range is `(sqrt(3)v^2)/(2g)`. The vertical range is `v^2/(8g)`. The angle which the projectile makes with the horizontal initially is:

Options

A. `15^@`
B. `30^@`
C. `45^@`
D. `60^@`

Answer

B

Explanation

Let be the angle of projection with the horizontal .
`:. (v^2 sin 2 theta)/g = sqrt(3v^2)/(2g)`
or `sin 2theta=sqrt3/2" or " 2 theta= 60^(@) " or " theta= 30^(@)`
For vertical range ,
`(v^2 sin^2 theta)/(2g)=v^2/(8g) " or " sin^(2) theta=1/4`
or `sin theta=1/2 " or " theta=30^@`