Question

Two parallel straight lines are inclined to the horizon at an angle $\alpha$. A particle is projected from a point mid way between them so as to graze one of the lines and strikes the other at right angle. Show that if $\theta$ is the angle between the direction of projection and either of lines, then $\tan \theta =(\sqrt{2}-1)\cot \alpha$.

Answer 1

$A$ to $B:0^2=u^2{\sin }^2-2\left(g\cos \alpha \right)h$
$A$ to $C:-h=u\sin \theta t-\frac{1}{2}g\cos \alpha t^2$
And $0=u\cos \theta -g\sin \alpha t\Rightarrow t=\frac{u\cos \theta }{g\sin \alpha }$
$-h=\frac{u^2\sin \theta \cos \theta }{g\sin \alpha }-\frac{1}{2}\frac{g\cos \alpha u^2{\cos }^2\theta }{g^2{\sin }^2\alpha }$
$\frac{-u^2{\sin }^2\theta }{2g\cos \alpha }=\frac{u^2\sin \theta \cos \theta }{g\sin \alpha }-\frac{u^2\cos \alpha {\cos }^2\theta }{2g{\sin }^2\alpha }$
$\frac{{\sin }^2\theta }{2\cos \alpha }=\frac{\sin \theta \cos \theta }{\sin \alpha }-\frac{\cos \alpha {\cos }^2\theta }{2{\sin }^2\alpha }$
Multiplying both sides by $\frac{2\cos \alpha }{{\cos }^2\theta }$, we get
$-{\tan }^2\theta =2\cot \alpha \tan \theta -{\cot }^2\alpha =0$
${\tan }^2\theta +2\cot \alpha \tan \theta -{\cot }^2\alpha =0$
$\tan \theta \frac{-2\cot \alpha +\sqrt{4{\cot }^2\alpha +4{\cot }^2\alpha }}{2}$
$\tan \theta =(\sqrt{2}-1)\cot \alpha$.