Question

A cannon fired from under a shelter inclined at an angle $\alpha$ to the horizontal. The cannon is at point A distant L from the base (B) of the shelter. The initial velocity of the cannon is $V_0$ and its trajectory lies in the plane. The maximum range $R_{max}$ of the shell is then

• A. $\frac{V_0^2}{g}\sin 2\alpha$
• B. $\frac{g}{v_0^2}\sin 2(\varphi -\alpha )$
• C. $\frac{V_0^2}{g}\sin 2$ $(\alpha +{\sin }^{-1}\frac{\sqrt{gR\sin 2\alpha }}{v_0})$
• D. $\frac{V_0^2}{g}\sin 2$ $(\alpha +{\sin }^{-1}\frac{\sqrt{R\sin 2\alpha }}{g})$

Answer 1

The correct option is C $\frac{V_0^2}{g}\sin 2$ $(\alpha +{\sin }^{-1}\frac{\sqrt{gR\sin 2\alpha }}{v_0})$

$Horigontalcomponentofv=v\cos (\beta -\alpha )$

$Verticalcomponentofv=v\sin (\beta -\alpha )$

$h=CD=AB\to$

$\frac{AB}{l}=\sin \alpha$ $\&$

$AB=l\sin \alpha$ $or$

$h=l\sin \alpha$

$h=\frac{v^2{\sin }^2(\beta -\alpha )}{2g^{\prime }}-1$

$l\sin \alpha =\frac{v^2{\sin }^2(\beta -\alpha )}{2g\cos \alpha }$

$v^2{\sin }^2(\beta -\alpha )=2\lg \sin \alpha \cos \alpha$

$v^2{\sin }^2(\beta -\alpha )=\lg \sin \left(2\alpha \right)$

$v^2\le gl\sin 2\alpha ;$

$\beta =90$

$R_{max.}=\frac{v^2}{g}$

$v^2>gl\sin 2\alpha$

$Max.range\to$

$v^2{\sin }^2(\beta -\alpha )=\lg \sin \left(2\alpha \right)$

$R_{max.}=\frac{{v_0}^2}{g}\sin 2\left(\alpha +{\sin }^{-1}\frac{\sqrt{gR\sin 2\alpha }}{v}\right)$