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 horizontal motion of the projectile, giving it a constant horizontal acceleration equal to $g$. Under the same conditions of projection, the new range will be (g = acceleration due to gravity)

• A. $R+H$
• B. $R+2H$
• C. $R+\frac{3H}{2}$
• D. R + 4H

Answer 1

Answer: (B)

$R+2H$

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 $R_1=u_{x}T+\frac{1}{2}a{T}^2$

where $T=\frac{2u\sin \theta }{g}$ is the time of flight.

Or

$R_1=\left(u\cos \theta \right)\frac{2u\sin \theta }{g}+\frac{g}{4}\times (\frac{2u\sin \theta }{g}{)}^2$

Or

$R_1=\frac{u^2\sin 2\theta }{g}+2\frac{u^2{\sin }^2\theta }{2g}$

Using $R=\frac{u^2\sin 2\theta }{g}$ and $H=\frac{u^2{\sin }^2\theta }{2g}$

$⟹R_1=R+2H$