Question

A projectile is projected with speed $u$ at an angle of ${60}^o$ with horizontal from the foot of an inclined plane. If the projectile hits the inclined plane horizontally, the range on inclined plane will be

• A. $\frac{u^2\sqrt{21}}{2g}$
• B. $\frac{3u^2}{4g}$
• C. $\frac{u^2}{2\beta }$
• D. $\frac{\sqrt{21}{u}^2}{8g}$

Answer 1

The correct option is D $\frac{\sqrt{21}{u}^2}{8g}$

Since the projectile hits the inclined plane horizontally, this point would have been the highest point for the projectile if the inclined plane were not there.
$\therefore$ horizontal distance of the point from the point of projection to where the projectile hits the inclined plane is $\frac{R}{2}=\frac{u^{2}sin2\left(\theta \right)}{2g}=\frac{u^{2}sin2\left({60}^{\circ }\right)}{2g}=\frac{\sqrt{3}{u}^2}{4g}$

$H=\frac{u^{2}si{n}^2\theta }{2g}$

$H=\frac{u^{2}si{n}^2{60}^{\circ }}{2g}=\frac{3u^2}{8g}$

$\therefore AB=\sqrt{{\left(\frac{R}{2}\right)}^2+H^2}=\frac{\sqrt{21}{u}^2}{8g}$