Question

A ball is projected horizontally relative to the plane with a speed $v$ from the top of a plane inclined at an angle ${45}^0$ with the horizontal. How far from the point of projection will the ball strike the plane?

• A. $\frac{v^2}{g}$
• B. $\sqrt{2}\frac{v^2}{g}$
• C. $\frac{2v^2}{g}$
• D. $\frac{2\sqrt{2}{v}^2}{g}$

Answer 1

The correct option is D $\frac{2\sqrt{2}{v}^2}{g}$

Writing equation of motion in normal to incline plane:

$S_y=vsin\alpha t-\frac{gcos\alpha t^2}{2}$

For Range , $S_y=0$
which gives $t=\frac{2vsin\alpha }{gcos\alpha }$
$S_x=vcos\alpha t+\frac{gsin\alpha t^2}{2}$

Replacing the value of $t$, writing Equation of motion in along incline plane

$S_x=\left(vcos\alpha \right)t+\left(\frac{gsin\alpha }{2}\right)\left(t^2\right)$

$S_x=\left(vcos\alpha \right)\left(\frac{2Vsin\alpha }{gcos\alpha }\right)+\left(\frac{gsin\alpha }{2}\right)\left({\frac{2Vsin\alpha }{gcos\beta }}^2\right)$

$S_x=\frac{2v^{2}sin\alpha }{gcos\alpha }(cos\alpha +\frac{si{n}^2\alpha }{cos\alpha })$

$S_x=\frac{2v^2}{g}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})=\frac{2\sqrt{2}{v}^2}{g}$