Question

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

• A. $R+H$
• B. $R+5H$
• C. $R+2H$
• D. $R+3H$

Answer 1

Answer: (C)

$R+2H$

Let the angle of projection be $\theta$ and speed of projectile be $u$.
Initially, horizontal component of velocity $u_x=u\cos \theta$
Acceleration in horizontal direction $a=\frac{g}{2}$
So, horizontal range of projectile $R^{\prime }=u_{x}T+\frac{1}{2}a{T}^2$
where $T=\frac{2u\sin \theta }{g}$ is the time of flight.
Or $R^{\prime }=\left(u\cos \theta \right)\frac{2u\sin \theta }{g}+\frac{g}{4}\times (\frac{2u\sin \theta }{g}{)}^2$
Or $R^{\prime }=\frac{u^2\sin 2\theta }{g}+2\frac{u^2{\sin }^2\theta }{2g}$
Using $R=\frac{u^2\sin 2\theta }{g}$ and $H=\frac{u^2{\sin }^2\theta }{2g}$
$⟹R^{\prime }=R+2H$