A bullet fired from a gun falls at a distance half of its maximum range. The angle of projection of the bullet is

Options

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

Answer

D

Explanation

As, we know that for a projectile motion, maximum range,
`R_("max") = (u^(2))/(g)` and `h_("max") = (R_("max"))/(4)`
Given, `R = (R_("max"))/(2)`
`rArr (R_("max"))/(2) = (u^(2))/(2g) = R` ….(i)
where, `R = (u^(2) sin 2 theta)/(g)`
Hence, `(u^(2) sin 2 theta)/(g) = (u^(2))/(2g)` [From Eq. (i)]
`rArr sin 2 theta = (1)/(2)`
`rArr 2 theta = 30^(@) rArr theta 15^(@)`
Hence, the correct option is (d).