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

Let u: velocity of projection and : angle of projection `rArr R=(u^(2) sin 2theta)/(g) and H=(u^(2) sin^(2) theta)/(2g)`
If horizontal acceleration = g/4 is imparted: Let T : time of Flight
`S_(y)=0 rArr 0=u sin theta T-1/2gT^(2) rArr T=(2u sin theta)/(g)`
New range: `S_(x)=u cos thetaT +1/2 g/4 T^(2)=(2u^(2) sin theta cos theta)/(g)+1/2 g/4. (4u^(2) sin^(2) theta)/(g^(2))= R + H `
From initial point of projection to maximum height gain, `V_(y)=0`