Question

A shot is fixed from O at an angle $\alpha$ with horizontal. If the shot hits the object P on an inclined plane horizontally then the relation between $\alpha$ and $\beta$ is

• A. $tan\alpha =\frac{tan\beta }{2}$
• B. $tan\alpha =2tan\beta$
• C. $tan\alpha =\frac{tan\beta }{4}$
• D. $tan\alpha =4tan\beta$

Answer 1

The correct option is B $tan\alpha =2tan\beta$

Finally, the particle has horizontal speed only, Hence this is the top most point of the projectile

Hence, $H=\frac{u^2{\sin }^2\alpha }{2g}$

& $R=\frac{1}{2}\times \frac{2u^2\sin \alpha \cos \alpha }{g}$

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

From trigonometics

$\frac{u}{R}=\tan \beta$

$\frac{\sin \alpha }{2\cos \beta }=\tan \beta$

or, $\beta ={\tan }^{-1}\left(\frac{\sin \alpha }{2\cos \beta }\right)$

or, $\tan \alpha =2\tan \beta$