Question

A cannon on the level plane is aimed at an angle $\theta$ above the horizontal and a shell is fired with a muzzle velocity $v_o$ towards the vertical cliff at a distance $R$ away. The height from the bottom at which the shell strikes the side walls of the cliff is

• A. $R\tan \theta -\frac{g{R}^2}{2{v_o}^2{\cos }^2\theta }$
• B. $R\tan \theta -\frac{g{R}^2}{2{v_o}^2}$
• C. $R\tan \theta -\frac{g{R}^2}{2{v_o}^2{\sin }^2\theta }$
• D. $R\tan \theta +\frac{g{R}^2}{2{v_o}^2{\cos }^2\theta }$

Answer 1

The correct option is A $R\tan \theta -\frac{g{R}^2}{2{v_o}^2{\cos }^2\theta }$

For horizontal direction $u=vCos\theta$, acceleration is zero so x(t) $=vCos\theta .t$ and for vertical direction $u=vSin\theta$, acceleration is g so $y\left(t\right)=vSin\theta .t-\frac{1}{2}g.t^2$

Eleminating t from both equations.

$Y\left(t\right)=vSin\theta \times \frac{x\left(t\right)}{vCos\theta }-\frac{1}{2}g(\frac{x\left(t\right)}{vCos\theta }{)}^2$ where $\theta$,here $v=v_0$ and x(t)=R, y(t)=H.

So $H=Rtan\theta -\frac{g{R}^2}{2v_0^{2}Co{s}^2\theta }$

So optionA is best possible amswer.