Question

An inclined plane making an angle $\beta $ with horizontal. A projectile projected from the bottom of the plane with a speed u at angle $\alpha $ with horizontal, then its maximum range $R_{m a x}$ is

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

Answer 1

Answer: (B)

The expression of horizontal Range $R=\frac{2 u^{2} sin \left(\alpha - \beta \right) cos ⁡ \alpha }{g c o s^{2} \beta }$
From identity
$2sin Acos ⁡ B=sin ⁡ \left(A + B\right) + sin ⁡ \left(A - B\right)$
$2sin \left(\alpha - \beta \right) cos ⁡ \alpha =sin ⁡ \left(2 \alpha - \beta \right)+sin ⁡ \left(- \beta \right)$
$R=\frac{u^{2}}{g cos^{2} \beta }$ $\left[sin \left(2 \alpha - \beta \right) - sin ⁡ \beta \right]$
Range R along incline is maximum if $2\alpha -\beta =\frac{\pi }{2}$ (as β is fixed)
$\text { So, } R_{\max }=\frac{u^2}{g \cos ^2 \beta}[1-\sin \beta] $
$=\frac{u^2}{g\left(1-\sin ^2 \beta\right)}[1-\sin \beta] $
$ =\frac{u^2}{g(1+\sin \beta)(1-\sin \beta)}[1-\sin \beta] $
$ =\frac{u^2}{g(1+\sin \beta)}$