Question

An inclined plane is making an angle $\beta$ with horizontal. A projectile is projected from the bottom of the plane with a speed $u$ at an angle $\alpha$ with horizontal then its range $R$ when it strikes the inclined plane is :

• A. $\frac{2u^{2}sin(\alpha -\beta )cos\alpha }{gco{s}^2\beta }$
• B. $\frac{u^{2}sin(\alpha -\beta )cos\alpha }{gco{s}^2\beta }$
• C. $\frac{2u^{2}sin(\alpha +\beta )cos\alpha }{gco{s}^2\beta }$
• D. $\frac{u^{2}sin(\alpha +\beta )cos\alpha }{gco{s}^2\beta }$

Answer 1

The correct option is A $\frac{2u^{2}sin(\alpha -\beta )cos\alpha }{gco{s}^2\beta }$

To calculate time of flight:

We know that the total displacement perpendicular to the inclined plane is zero.
So, $0=u\sin (\alpha -\beta )t-\frac{1}{2}g\cos \beta t^2$
$\therefore t=\frac{2usin(\alpha -\beta )}{gcos\beta }$
Now, for range:
$OB=\left(ucos\alpha \right)t=\frac{2u^{2}sin(\alpha -\beta )cos\alpha }{gcos\beta }$
$\therefore OA=OB/cos\beta =\frac{2u^{2}sin(\alpha -\beta )cos\alpha }{gco{s}^2\beta }=R$