Question

A particle is projected at an angle $\alpha$ with the horizontal from the foot of a plane, whose inclination to the horizontal is $\beta$. Find the velocity with which the particle strikes perpendicular to the inclined plane.

• A. $u\cos (\alpha -\beta )$
• B. $u\sin (\alpha -\beta )$
• C. $u\sin \left(\frac{\alpha }{2}\right)$
• D. $u\cos \left(\frac{\beta }{2}\right)$

Answer 1

The correct option is B $u\sin (\alpha -\beta )$


As the final velocity $v$ is along Y-direction, we can say
$v=u\sin (\alpha -\beta )-g\cos \beta T$ where $T$ is the time of flight.
$T=\frac{2u\sin (\alpha -\beta )}{g\cos \beta }$
$\Rightarrow v=u\sin (\alpha -\beta )-g\cos \beta \left[\frac{2u\sin (\alpha -\beta )}{g\cos \beta }\right]\therefore v=-u\sin (\alpha -\beta )$
Thus, velocity with which particle strikes the plane is $u\sin (\alpha -\beta )$