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

`H' = H` (as vertical component of acceleration has not changed)
`R' = u_(x) T +(1)/(2)a_(x) T^(2) = R +(1)/(2) xx (g)/(4) xx (4u^(2)sin^(2) theta)/(g^(2))`
`= R + (u^(2)sin^(2) theta)/(2g) = (R +H)`