Question

A gunman sitting in a shelter fires a bullet. The shelter is at an angle $'\alpha '$ to the horizontal. The gunman is at point A, which is at a distance of \[100meters\] from the base of the shelter as shown in the figure. The initial velocity of the gun shot is \[{V_a}\] ​ and its trajectory lies in the plane of the figure. Determine the maximum range of gunshot.
 

Answer 1

Hint:In this question, it is given that the trajectory of the gun shot lies in the plane of the figure. However, we will consider the trajectory that touches the shelter from all the possible trajectories. By doing this, the angle at which the bullet should be fired to touch the shelter. And finally by the help of this angle, we will determine the maximum range of the gunshot.

Complete step by step answer:
We will first derive the formula for the maximum range of the gunshot and then put the value of the base given in the figure. Let the base be $l$ and consider the following figure for the trajectory where the bullet touches the shelter.

We can determine from the figure that the horizontal component of the initial velocity of the bullet in this system is \[{v_{0x}} = {v_0}\cos \left( {\theta {\text{ }} - \;\alpha } \right)\], and the vertical component is \[{v_{0y}} = {v_0}\sin \left( {\theta {\text{ }} - \;\alpha } \right)\], where $\theta $ is the angle formed by the direction of the initial velocity of the shell and the horizontal.

Point C at which the trajectory of the bullet touches the shelter gives us the maximum altitude of the shell above the horizontal which is given by:
$h' = \dfrac{{{v_{0y}}^2}}{{2g'}}$, where $g' = g\cos \alpha $ which is the free fall acceleration for the considered coordinate system.
We can thus write that
${v^2}{\sin ^2}\left( {\theta - \alpha } \right) = 2gl\cos \alpha \sin \alpha \\
\Rightarrow \theta = \alpha + {\sin ^{ - 1}}\left( {\dfrac{{\sqrt {gl\sin 2\alpha } }}{{{v_0}}}} \right) \\ $
Now, the maximum range of the gunshot can be given by:
${R_{\max }} = \dfrac{{{v_0}^2}}{g}\sin 2\theta \\
\Rightarrow {R_{\max }} = \dfrac{{{v_0}^2}}{g}\sin 2\left( {\alpha + {{\sin }^{ - 1}}\left( {\dfrac{{\sqrt {gl\sin 2\alpha } }}{{{v_0}}}} \right)} \right) \\ $
Now we will put $l = 100m$ as given in the figure.
$\therefore{R_{\max }} = \dfrac{{{v_0}^2}}{g}\sin 2\left( {\alpha + {{\sin }^{ - 1}}\left( {\dfrac{{\sqrt {100g\sin 2\alpha } }}{{{v_0}}}} \right)} \right)$

Hence,the maximum range of gunshot is $\dfrac{{{v_0}^2}}{g}\sin 2\left( {\alpha + {{\sin }^{ - 1}}\left( {\dfrac{{\sqrt {100g\sin 2\alpha } }}{{{v_0}}}} \right)} \right)$.

Note:In this question, we have used the coordination system for the trajectory when the bullet touches the shelter. We have considered the angle of projection in this trajectory only. Also, the maximum amplitude of the gunshot is considered in this trajectory. Therefore, we should not be confused between the original and the considered coordination system.