Question

Two inclined planes intersect in a horizontal plane. their inclinations to the horizontal being $\alpha$ and $\beta$. If a particle is projected with velocity u at right angle to the former from a point on it, find the time after which the velocity vector will become perpendicular to the other inclined plane.

• A. $t=\frac{usin(\alpha +\beta )}{gsin\beta }$
• B. $t=\frac{ucos(\alpha +\beta )}{gcos\beta }$
• C. $t=\frac{ucos(\alpha +\beta )}{gsin\beta }$
• D. $t=\frac{usin(\alpha +\beta )}{gcos\beta }$

Answer 1

The correct option is B $t=\frac{ucos(\alpha +\beta )}{gcos\beta }$

$v\cos \beta =u\sin \alpha$

$v\sin \beta =u\cos \alpha -gt$

$\Rightarrow \frac{u\sin \alpha }{\cos \beta }\cdot \sin \beta =u\cos \alpha -gt$

$\Rightarrow \frac{u\sin \alpha \sin \beta +u\cos \alpha \cos \beta }{\cos \beta }=gt$

$\Rightarrow t=\frac{u\cos \left(\alpha +\beta \right)}{g\cos \beta }$

Hence,

option $\left(B\right)$ is correct answer.