Question

A particle is projected at an angle $\alpha$ with the horizontal from the foot of a plane, whose inclination to the horizontal is $\beta$. What is the value of $\tan (\alpha -\beta )$ if it strikes the plane at right angle?

• A. $\frac{\cot \beta }{2}$
• B. $\frac{\sin \beta }{2}$
• C. $\frac{\tan \beta }{2}$
• D. $\text{cosec}\beta$

Answer 1

The correct option is A $\frac{\cot \beta }{2}$


Let us analyze the values of initial velocity, final velocity and acceleration along X and Y axis
$$\begin{array}{cc}\text{X - axis}& \text{Y- axis}\\ u_x=u\cos (\alpha -\beta )& u_y=u\sin (\alpha -\beta )\\ v_x=0& v_y=?\\ a_x=-g\sin \beta & a_y=-g\cos \beta \end{array}$$
If final velocity $v$ is perpendicular to plane,
Time of flight $T=\frac{2u\sin (\alpha -\beta )}{g\cos \beta }$
Also, using first equation of motion,
$v_x=u_x+at=u\cos (\alpha -\beta )-g\sin \beta \times T=0$

$\Rightarrow u\cos (\alpha -\beta )=\left(g\sin \beta \right)\frac{2u\sin (\alpha -\beta )}{g\cos \beta }\Rightarrow \cos \beta =2\tan (\alpha -\beta )\sin \beta \therefore \tan (\alpha -\beta )=\frac{\cot \beta }{2}$