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 $R_{max}$ is

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

Answer 1

Answer: (B)

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