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

`T=(2u_(y))/g, H=(u_(y)^(2))/(2g)`
and `R=u_(x)T`
When a horizontal acceleration is also given to the projectile `u_(y)` and T and H will remain unchanged while the range will become
`R'=u_(x)T+1/2aT^(2)`
`=R+1/2 g/4((4u_(y)^(2))/(g^(2)))=R+H`.