Question

A ball is projected horizontally with a speed $v$ from the top of a plane inclined at an angle ${45}^{\circ }$ 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. $\sqrt{2}\left[\frac{2v^2}{g}\right]$

Answer 1

The correct option is D $\sqrt{2}\left[\frac{2v^2}{g}\right]$

Let the distance along inclined plane be $l$ at which the ball will strike.
Let the horizontal distance covered by the ball before striking be x.
$\begin{array}{}\Rightarrow x=vt& \text{(1)}\end{array}$
Vertical distance covered by the ball is given by
$\begin{array}{}y=\frac{1}{2}g{t}^2& \text{(2)}\end{array}$
Using value of $t$ from (1) in (2), we get
$y=\frac{g{x}^2}{2v^2}$
From the figure, $\tan {45}^{\circ }=\frac{y}{x}$
or $y=xx=\frac{g{x}^2}{2v^2}\Rightarrow x=\frac{2v^2}{g}\Rightarrow y=\frac{2v^2}{g}$
From the figure, we get,
$l^2=x^2+y^{2}l=\sqrt{2}\left[\frac{2v^2}{g}\right]$