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 $0=u\cos \theta -g\sin \alpha t$ $\Rightarrow t=\frac{u\cos \theta }{g\sin \alpha }$

$-h=\frac{u^2\sin \theta \cos \theta }{g\sin \alpha }-\frac{1g\cos \alpha u^2{\cos }^2\theta }{g^2{\sin }^2\alpha }$

$\frac{-u^2{\sin }^2\theta }{2g\cos \alpha }=\frac{u^2\sin \theta \cos \theta }{g\sin \alpha }-\frac{u^2\cos \alpha {\cos }^2\theta }{2g{\sin }^2\alpha }$

$\frac{-{\sin }^2\theta }{2\cos \alpha }=\frac{\sin \theta \cos \theta }{\sin \alpha }-\frac{\cos \alpha {\cos }^2\theta }{2{\sin }^2\alpha }$

Multiplying both sides by $\frac{2\cos \alpha }{{\cos }^2\alpha }$ we get

$-{\tan }^2\theta =2\cot \alpha \tan \theta -{\cot }^2\alpha =0$

${\tan }^2\theta +2\cot \alpha \tan \theta -{\cot }^2\alpha =0$

$\tan \theta =\frac{-2\cot \alpha +\sqrt{4{\cot }^2\alpha +4{\cot }^2\alpha }}{2}$

$=(\sqrt{2}-1)\cot \alpha$