An enemy ship is on the east side of a mountainous island. The enemy ship can maneuver to within 2500 meters of the 1800 meter high mountain peak and can shoot projectiles with an initial speed of 250m/sec. If western shoreline is horizontally 300 meters 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?

Draw the figure. Figure the max range is at a shooting angle of 45 deg.

Determine initial vertical vertical velocity (250)sin45

Deterine the time in air:
finalvertical position = initial vertical position + initialverticalvelocity*time + 1/2 (g)time^2
the initial and final vertical postion is zero, so

250sin45t - 4.9 t^2=0
solve for time in air.
Now, your horizontal equation.
xfinal = Vxi *time

solve for x final.Subtract 300+2500 and you have it. Check my thinking.

To solve this problem, we can use the equations of projectile motion.

First, let's draw a diagram. The mountain peak is located on the eastern side of the island, and the western shoreline is horizontally 300 meters away from the peak. The enemy ship can maneuver to within 2500 meters of the peak.

The maximum range is achieved when the projectile is launched at a 45-degree angle. This is because the vertical and horizontal components of the initial velocity are equal, resulting in the longest possible distance traveled.

The initial vertical velocity can be determined using the equation: initial vertical velocity = initial velocity * sin(angle). In this case, the initial velocity is given as 250 m/sec, and the angle is 45 degrees. So, the initial vertical velocity is 250 * sin(45) = 176.78 m/sec.

Next, let's determine the time in the air. We can use the equation: final vertical position = initial vertical position + initial vertical velocity * time + (1/2) * g * time^2. Since the initial and final vertical positions are both zero (as the object travels in a parabolic trajectory), the equation becomes: 0 = 176.78t - (4.9)t^2, where g is the acceleration due to gravity (approximately 9.8 m/sec^2).

Solving this equation will give us the time it takes for the projectile to reach the peak and fall back to the ground. Let's solve it:

0 = 176.78t - (4.9)t^2
4.9t^2 - 176.78t = 0
t(4.9t - 176.78) = 0

We have two possible solutions: t = 0 (which is not of interest to us) and t = 36.03 seconds.

Now, let's determine the horizontal distance covered by the projectile. We can use the equation: final horizontal position = initial horizontal velocity * time. Since the initial horizontal velocity is the same as the initial velocity (250 m/sec) and the time in the air is 36.03 seconds, we can calculate the horizontal distance:

final horizontal position = 250 * 36.03 = 9007.5 meters.

Finally, we subtract the distance from the peak to the western shoreline (300 meters) and the distance from the peak to within the range of the enemy ship (2500 meters) from the final horizontal position to find the safe distances from the western shore where a ship can be safe from bombardment:

Safe distance = 9007.5 - 300 - 2500 = 6207.5 meters.

So, ships located at a horizontal distance greater than 6207.5 meters from the western shore will be safe from the bombardment of the enemy ship.