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 [tex]v_{0}[/tex] 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
[tex]v_{x}=v_{0} \times \cos \theta[/tex]
As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of [tex]v_{x}[/tex] and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by
[tex]v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}[/tex]
[tex]v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}[/tex]
For motion with constant acceleration, we know
[tex]s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}[/tex]
Along the horizontal, x-axis, we might write this as
[tex]x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}[/tex]
Measuring distances relative to the firing point means
[tex]x_{0}=0[/tex]
we know that,
[tex]a_{x}=0[/tex]
or,
[tex]v_{x}=v_{x 0}=\text { constant }[/tex]
By applying the values, we get,
[tex]x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}[/tex]
The acceleration of gravity is vertically downward and is [tex]g=-9.8 \mathrm{m} / \mathrm{s}^{2}[/tex] , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion
[tex]y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}[/tex]
we know, [tex]y_{0}=0[/tex] and [tex]a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}[/tex], so,
[tex]y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)[/tex]
y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m
