Question

A particle is projected up an inclined plane of inclination $\beta$ at an elevation a to the horizon. Show that
a. $\tan \alpha +\cot \beta +2\tan \beta$, if the particle strikes the plane at right angles
b. $\tan \alpha =2\tan \beta$ if the particle strikes the plane horizontally.

Answer 1

a. $0=u\sin (\alpha -\beta )t-\frac{1}{2}g\cos \beta t^2$
$t=2u\sin (\alpha -\beta )/g\cos \beta ....\left(i\right)$
$0=u\cos (\alpha -\beta )-g\sin \beta t$
$\Rightarrow t=\frac{u\cos (\alpha -\beta )}{g\sin \beta }....\left(ii\right)$
From (i) and (ii), we get
$\frac{2\sin (\alpha -\beta )}{\cos \beta }=\frac{\cos (\alpha -\beta )}{\sin \beta }\Rightarrow 2\tan (\alpha -\beta )=\cot \beta$
$\Rightarrow \frac{2(\tan \alpha -\tan \beta )}{1+\tan \alpha \tan \beta }=\cot \beta$
$\Rightarrow 2\tan \alpha -2\tan \beta =\cot \beta +\tan \alpha$
$\Rightarrow \tan \alpha =\cot \beta +2\tan \beta$
b. $t=2u\sin (\alpha -\beta )/g\cot \beta$
Also, $t=\frac{u\sin \alpha }{g}\Rightarrow \frac{2\sin (\alpha -\beta )}{\cos \beta }=\sin \alpha$
$\Rightarrow 2\sin \alpha \cos \beta -2\cos \alpha \sin \beta =\sin \alpha \cos \beta$
$\Rightarrow \sin \alpha \cos \beta =2\cos \alpha \sin \beta$
$\Rightarrow \tan \alpha =2\tan \beta$.