The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to below in the direction of motion of the projectile, giving it a constant horizontal acceleration `=g//2`. Under the same conditions of projection. Find the horizontal range of the projectile.

Options

A. `R+H`
B. `R+2H`
C. `R`
D. `R+H//2`

Answer

B

Explanation

New horizontal range
`=R+1/2xxg/2xxT^(2)=R+g/4xx(4u^(2)sin^(2)theta)/g^(2)`
`R+2H ( :' H=(u^(2) sin^(2) theta)/(2g))`