Question

An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explodes on a mountainside 49.5 s after firing. What are the x- and y-coordinates of the shell where it explodes, relative to its firing point?

Answer 1

The x- and y-coordinates are 9142.57 m and -304.425 m

Explanation:

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell $v_{0}$ in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

                $v_{x}=v_{0} \times \cos \theta$

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of $v_{x}$ and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

           $v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}$

           $v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}$

For motion with constant acceleration, we know

            $s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}$

Along the horizontal, x-axis, we might write this as

            $x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}$

Measuring distances relative to the firing point means

               $x_{0}=0$

we know that,

              $a_{x}=0$

or,

             $v_{x}=v_{x 0}=\text { constant }$

By applying the values, we get,

           $x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}$

The acceleration of gravity is vertically downward and is $g=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

           $y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}$

we know, $y_{0}=0$ and $a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}$, so,

          $y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)$

                 y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m