Answer:
The shell will land 10.18m away from the buoy.
Explanation:
In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).
Now, there are two cases, one with the tailwind and another with the tailwind. In both cases the shell would have the same vertical initial velocity and acceleration, therefore the shell would hit the water in the same amount of time. So we need to first find the time it takes the shell to hit the water.
In order to do so we can use the following equation:
[tex]y_{f}=y_{0}+V_{0}t+\frac{1}{2}at^{2}[/tex]
now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:
[tex]0=y_{0}+\frac{1}{2}at^{2}[/tex]
and solve for t:
[tex]t=\sqrt{\frac{-2y_0}{a}}[/tex]
now we can substitute the values:
[tex]t=\sqrt{\frac{-2(50m)}{-9.81m/t^2}}[/tex]
t=3.19s
Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.
So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the distance from the ship to point A to be:
[tex]x_{A}=V_{x}t[/tex]
[tex]x_{A}=(100m/s)(3.19)[/tex]
[tex]x_{A}=319m[/tex]
We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:
[tex]x_{f}=V_{x0}t+\frac{1}{2}a_{x}t^{2}[/tex]
So we can substitute the given values:
[tex]x_{f}=(100m/s)(3.19s)+\frac{1}{2}(2m/s^{2})(3.19s)^{2}[/tex]
Which yields:
[tex]x_{f}=329.18m[/tex]
so now we can use the A and B points to find by how far the shell missed the buoy:
Distance=329.18m-319m=10.18m
So the shell missed the buoy by 10.18m.
