Question

A cannon on a level plain is aimed at an angle $\alpha$ above the horizontal and a shell is fired with a muzzle velocity $v_0$ towards a vertical cliff at a distance R away. Then the height from the bottom at which the shell strikes the side walls of the cliff is

• A. $Rsin\alpha -\frac{g{R}^2}{2v_0^{2}si{n}^2\alpha }$
• B. $Rcos\alpha -\frac{g{R}^2}{2v_0^{2}co{s}^2\alpha }$
• C. $Rtan\alpha -\frac{g{R}^2}{2v_0^{2}co{s}^2\alpha }$
• D. $Rtan\alpha -\frac{g{R}^2}{2v_0^{2}si{n}^2\alpha }$

Answer 1

The correct option is C

$Rtan\alpha -\frac{g{R}^2}{2v_0^{2}co{s}^2\alpha }$

The time taken to move a horizontal distance R is $t=\frac{R}{v_{0}cos\alpha }$.
Therefore, the vertical distance moved in this time is given by
$h=v_{0}sin\alpha t-\frac{1}{2}g{t}^2=v_{0}sin\alpha \times \frac{R}{v_{0}cos\alpha }-\frac{1}{2}g{\left(\frac{R}{v_{0}cos\alpha }\right)}^2=Rtan\alpha -\frac{g{R}^2}{2v_0^{2}co{s}^2\alpha }$
Hence, the correct choice is (c).