In the absence of wind the range and maximum height of a projectile were R and H. If wind imparts a horizontal acceleration a = g/4 to the projectile then find the maximum range and maximum height.

Explanation

`H^(1) =H` (`therefore u sin theta` remains same)
`T.=T`
`R. =u_(x) T + 1/2 aT^(2) =R + 1/2 g/4 T^(2)`
`=R + 1/8 gT^(2) =R + H`
`R^(1) =R + H`
`H^(1)=H`