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 maximum range $R_{max}$ is

• A. $R_{max}=\frac{u^2}{g(1-sin\beta )}$
• B. $R_{max}=\frac{u^2}{g(1+sin\beta )}$
• C. $R_{max}=\frac{u}{g(1-sin\beta )}$
• D. $R_{max}=\frac{u}{g(1+sin\beta )}$

Answer 1

The correct option is B $R_{max}=\frac{u^2}{g(1+sin\beta )}$

The expression of horizontal range $R=\frac{2u^2\sin (\alpha -\beta )\cos \alpha }{g{\cos }^2\beta }$

From Identity $2\sin A\cos B=\sin (A+B)+\sin (A-B)$

$2\sin (\alpha -\beta )\cos \beta =\sin (2\alpha -\beta )+\sin (-\beta )$

$R=\frac{u^2}{g{\cos }^2\beta }\left[\sin \right(2\alpha -\beta )-\sin \beta ]$

Range R along incline is maximum if $2\alpha -\beta =\pi /2$

so, $R_{max}=\frac{u^2}{g{\cos }^2\beta }[1-\sin \beta ]$

$=\frac{u^2}{g(1-si{n}^2\beta )}[1-\sin \beta )]$

$=\frac{u^2}{g(1+sin\beta )(1-sin\beta )}[1-\sin \beta ]=\frac{u^2}{g(1+sin\beta )}$