The horizontal range and miximum height attained by a projectile are `R and H`, respectively. If a constant horizontal acceleration `a = g//4` is imparted to the projectile due to wind, then its horizontal range and maximum height will be

Options

A. `(R + H),(H)/(2)`
B. `(R + (H)/(2)), 2H`
C. `(R + 2H),H`
D. `(R + H), H`

Answer

D

Explanation

(d) `H = (u^2 sin^2 theta)/(2 g), R = (u^2 sin 2 theta)/(g)`
Maximum height will be same bacause acceleration `a = g//4` is in horizontal direction
`R' = u cos theta T + (1)/(2) a T^2 = R + (1)/(2) (g)/(4) ((2u sin theta)/(g)) = R + H`.