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 $l$ be the distance along inclined plane at which the ball will strike.
Let the horizontal distance covered by the ball before striking be $x$.
$\Rightarrow x=vt$ (i)

Vertical distance covered by the ball is given by
$y=\frac{1}{2}g{t}^2$ (ii)

Using value of $t$ from (i) in (ii), we get
$y=\frac{g{x}^2}{2v^2}$ (iii)

From the figure, $\tan {45}^{\circ }=\frac{y}{x}$
or $y=x$
$\Rightarrow x=\frac{g{x}^2}{2v^2}$ (from (iii))
$\Rightarrow x=\frac{2v^2}{g}$

So, $y=x=\frac{2v^2}{g}$

From the figure, we get, $l^2=x^2+y^2=\sqrt{2}\left[\frac{2v^2}{g}\right]$ is the distance along the inclined plane at which the ball will hit.

Or Alternative Method 1:


Applying second eqn of motion, in the direction perpendicular to the inclined plane,

$s=0$
i.e $v\cos {45}^{0}t-\frac{1}{2}g\sin {45}^0{t}^2=0$
$\Rightarrow t=\frac{2v}{g}$

Hence distance along the inclined plane
$x=v\sin {45}^{\circ }t+\frac{1}{2}g\sin {45}^{\circ }{t}^2=\frac{v}{\sqrt{2}}\times \frac{2v}{g}+\frac{1}{2}\times \frac{g}{\sqrt{2}}\times \frac{4v^2}{g}=\frac{2\sqrt{2}{v}^2}{g}$

Alternative Method 2 :

Since the ball is projected horizontally, horizontal distance covered
$x=v\sqrt{\frac{2y}{g}}$ and $x=y$ as angle is ${45}^{\circ }\Rightarrow \sqrt{x}=v\sqrt{\frac{2}{g}}\Rightarrow x=\frac{2v^2}{g}=y\therefore AB=x\sqrt{2}=2\sqrt{2}\frac{v^2}{g}$