Question

The horizontal range of a projectile is $R$ and the maximum height attained is $H$. A strong wind now begins to blow in the direction of motion of projectile giving it a horizontal acceleration equal to $\frac{g}{2}$. Under the same condition of projectile the horizontal range of projectile will be

• A. $R+\frac{H}{2}$
• B. $R+H$
• C. $R+\frac{3H}{2}$
• D. $R+2H$

Answer 1

The correct option is D $R+2H$

For projectile motion R$=\frac{u^2\sin 2\theta }{g}$
$H=\frac{u^2{\sin }^2\theta }{2g}$
Time of flight $t=\frac{2u\sin \theta }{g}$
Initial velocity $=u\cos \theta$
Horizontal acceleration due to wind$=\frac{g}{2}$
Horizontal range$=\left(u\cos \theta \right)t+\frac{1}{2}\left(\frac{g}{2}\right)t^2$
$=u\cos \theta .\frac{2u\sin \theta }{g}+\frac{g}{4}\frac{4u^2{\sin }^2\theta }{g^2}$
$=R+2H$