A ball is projected horizontally with a speed $v$ from the top of a plane inclined at an angle of 45 degrees with the horizontal. How far from the point of projection will the ball strike the plane?
My attempt at a solution:
I used the range formula for an inclined plane and I got the answer as $\sqrt{2} v^2/g$, which isn't the right answer. Any suggestions?
Horizontal component of velocity is constant (assumed), so $x=v\,t$ From acceleration due to gravity, $y=\frac{1}{2}\,g\,t^2$
The ball hits the plane when $x=y$ so set them equal to each other and solve for $t\, vt=\frac{1}{2}\,g\,t^2$: we get $v=\frac{1}{2}\,g\,t$ and $t=2\,\frac{v}{g}$.
At this time $t$, $x$ is equal to $y$.
To get the horizontal distance, plug in to the first equation:
$$x = v\times\left(2\,\frac{v}{g}\right)$$
so that $x = 2 v^2/g$
That is all
