Physics, 02.10.2019 01:20, Thin

Abattleship launches a shell horizontally at 100 m/s from the ship’s deck that’s 50 m above the water. the shell is intended to hit a target buoy in the water. however, when the shell is fired, a tailwind causes the shell to have a total acceleration of ~a = (2.0 m/s2 )ˆi − (9.81 m/s2 )ˆj. as a result, the shell misses its target. how far away from the buoy will the shell land? (to clarify, if there was no tailwind, the shell would hit the buoy.)

Answers: 1

Show answers

Answers

answered: bobby1415

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:

$y_{f}=y_{0}+V_{0}t+\frac{1}{2}at^{2}$

now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:

$0=y_{0}+\frac{1}{2}at^{2}$

and solve for t:

$t=\sqrt{\frac{-2y_0}{a}}$

now we can substitute the values:

$t=\sqrt{\frac{-2(50m)}{-9.81m/t^2}}$

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:

$x_{A}=V_{x}t$

$x_{A}=(100m/s)(3.19)$

$x_{A}=319m$

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:

$x_{f}=V_{x0}t+\frac{1}{2}a_{x}t^{2}$

So we can substitute the given values:

$x_{f}=(100m/s)(3.19s)+\frac{1}{2}(2m/s^{2})(3.19s)^{2}$

Which yields:

$x_{f}=329.18m$

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.

Abattleship launches a shell horizontally at 100 m/s from the ship’s deck that’s 50 m above the wate

answered: Guest

The answer to this is decibel

answered: Guest

An atom (ours, not carroll's) consists of a positively charged nucleus, surrounded by one or more negatively charged particles called electrons. the positive charges equal the negative charges, so the atom has no overall charge; it is electrically neutral. most of an atom's mass is in its nucleus; the mass of an electron is only 1/1836 the mass of the lightest nucleus, that of hydrogen. although the nucleus is heavy, it is quite small compared with the overall size of an atom. the radius of a typical atom is around 0.1 to 0.25 nanometres (nm), whereas the radius of a nucleus is about 10-6 nm. * if an atom were enlarged to the size of the earth, its nucleus would be only 200 feet in diameter and could easily rest inside a small football stadium.read more on -