Hard physics question on Projectile motion!!!?

Question

An enemy ship is on the east side of mountain island. The enemy ship has maneuvered to within 2500 m of the 1800 m high mountain peak and only can shoot projectiles with an initial speed of 250 m/s. If the western shoreline is horizontally 300 m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship?

Answer 1

I get that the maximum distance that the ship can fire is 6371m, meaning that any distance greater than 3571m from the western shore should be safe. It may be tempting to assume that the target ship would be safe within a certain minimum distance, where it would be protected by the mountain peak, but the enemy ship could hit any target within the 3571m limit by angling its guns higher than 45 degrees.

EDIT:
To start solving our problem, let's figure out the maximum distance that the enemy ship can fire. It's not hard, and we know we'll have to figure it out anyway.

If the ship fires its projectile at exactly 45 degrees, the projectile will make a parabolic arc through the air. At angles of more than 45 degrees, the projectile will fly higher, but it won't go as far (like a pop-up fly in baseball). At angles less than 45 degrees, the projectile will fly lower, and hit the water with more horizontal motion (like a line-drive in baseball). At exactly 45 degrees, it will achieve its maximum horizontal distance.

In order to figure out how far the projectile flies, we need to break up its motion into horizontal and vertical components. Imagine a right triangle with a hypotenuse of 250 units and one angle of 45 degrees. (The other angle will also be 45 degrees, but that doesn't matter for now.) The line adjacent to our angle will be as long as the projectile's horizontal momentum. The line opposite the angle will be as long as the projectile's vertical momentum. We just need a little bit of trigonometry to figure out these numbers.

sin(a) = opposite / hypotenuse;
sin(45) = Vv / 250; (Vv represents the vertical velocity)
Vv = sin(45) * 250 = 177m/s;

Since we're using a right triangle with two 45 degree angles, both sides will be the same length. If we were using a different angle, we could easily solve for the horizontal momentum using the Pythagorean Theorum.

So, we know that our projectile had a horizontal velocity of 177m/s and a vertical velocity of 177m/s. Let's figure out how long it would stay in the air. We know that it's constantly accelerating towards the earth due to the force of gravity. Our answer lies in the definition of acceleration: acceleration is a change in velocity over time. Mathematically, that's (a=ΔV / t). The "Δ" symbol is the Greek letter delta, often used to represent a change in a value.

To figure out how long it stayed up, we only need to worry about the projectile's vertical velocity. We know that it's initial vertical velocity is 177m/s. We also know that it's vertical velocity at the top of the arc is 0m/s. The change between these two points in time is (177m/s - 0m/s =) 177m/s. The only other thing we need to figure out time is the acceleration due to gravity. On Earth, it's always 9.81m/s^2. So...

a = ΔV / t;
9.81 = 177 / t;
t = 177 / 9.81 = 18.04 seconds. That's how long it takes the projectile to reach the top of it's arc. It will take equally as long to reach the bottom of it's arc, so the total flight time is 36.09 seconds.

Now we have a simple distance equation. Since distance is equal to velocity multiplied by time, the projectile's horizontal distance will be equal to its horizontal velocity multiplied by the amount of time that it stayed in the air.

d = 177 * 36 = 6372m. (This number is a little higher than the first one, because I rounded a little differently. It doesn't make a huge difference.) If we take out the distance to the western shore (2500m + 300M), we get a maximum distance of 3572m from the western shore. Any ship beyond that distance will be safe.

All we need to do now is figure out whether or not the enemy ship can fire at that maximum distance. If the mountain is too high, the ship will have to increase the angle of its guns, thereby losing range. We can figure out how high our projectile is going to fly with another simple distance equation. We've already determined that the projectile takes 18.04 seconds to reach the top of it's arc. We also know that it's moving with a vertical velocity of 177m/s. So...
distance = velocity * time;
distance = 177 * 18 = 3186m. Our projectile is plenty high enough to clear the top of the mountain.

So, the enemy ship can fire at a maximum distance of 3186m past the western shore and clear the top of the mountain with ease. For any targets closer than the maximum, all the ship needs to do is angle its guns higher.

Hope that helped.

Answer 2

I hope nobody uses this answer, because it's horribly wrong. The answerer forgot to take acceleration into account when determining whether or not the projectile could clear the mountain peak.

At a 45 degree angle, the shell fired from the enemy ship would only make it to 1520 meters high when out 2500m where the mountain peak is.