It was calculated that a shell when fired from a gun with a certain velocity and at an angle of elevation \[\dfrac{{5\pi }}{{36}}\] rad should strike a given target in the same horizontal plane. In actual practice, it was found that a hill just prevented the trajectory. At what angle of elevation should the gun be fired to hit the target.
A. $\dfrac{{5\pi }}{{36}}rad$
B. $\dfrac{{11\pi }}{{36}}rad$
C. $\dfrac{{7\pi }}{{36}}rad$
D. $\dfrac{{13\pi }}{{36}}rad$

Hint: Projectile motion is the basis for this enquiry. In order to hit the target accurately, we must determine the angle of elevation of the cannon in this question. Let's get a fundamental understanding of it first, and then we'll figure out how to solve it. The angle of elevation is the angle formed by a horizontal line and an oblique line from an object above the observer's eye.

Complete answer:
As for the question, a hill simply obstructed the path. In this situation, the shell should be fired at an angle of \[\left( {\dfrac{\pi }{2}{\text{ - }}\theta } \right)\] to hit the target.
Therefore, when a shell is fired from a gun at a specified velocity to impact a target, the angle of elevation is found to be \[\dfrac{{5\pi }}{{36}}rad\] . The trajectory was obstructed by a hill.
As a result, the needed angle of elevation for the gun to reach the target is:
\[\;\dfrac{\pi }{2} - \dfrac{{5\pi }}{{36}}{\text{ }} = \dfrac{{8\pi - 5\pi }}{{36}}rad = \dfrac{{13}}{{36}}rad\] .
As a result, the needed angle of elevation for firing the gun is $\dfrac{{13\pi }}{{36}}rad$

So, the correct option is: (D) $\dfrac{{13\pi }}{{36}}rad$

Note:
When addressing issues based on angles of elevation and depression, students should avoid becoming confused. The angle between the line of sight and the horizontal line is called angle of elevation when we perceive an object that is higher than the horizontal line. The angle between the line of sight and the horizontal line is called the angle of depression when we view an object below it.