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 ( :' u sin theta` remains same)
`T^(1)=T`
`R^(1)=u_(x)T+1/2aT^(2)=R+1/2g/4T^(2)`
`=R+1/8gT^(2),=R+H`
`R^(1)=R+H` ,`H^(1)=H`