An enemy ship is on the western side of a mountain island. The enemy ship can maneuver to within 2520 m of the 1850 m high mountain peak and can shoot projectiles with an initial speed of 260 m/s. If the eastern shoreline is horizontally 305 m from the peak, what is the minimum distance from the eastern shore at which a ship can be safe from the bombardment of the enemy ship?
What is the maximum distance from the eastern shore at which a ship can be safe from the bombardment of the enemy ship?
To calculate the minimum and maximum distances from the eastern shore at which a ship can be safe from the bombardment of the enemy ship, we need to consider the trajectory of the enemy's projectiles and the height of the mountain peak.
Let's start by visualizing the scenario. We have a mountain peak and a shoreline, with the enemy ship being on the western side of the mountain. The enemy ship can maneuver to within 2520 m of the mountain peak and shoot projectiles with an initial speed of 260 m/s.
To find the minimum distance from the eastern shore at which a ship can be safe, we need to consider the trajectory of the projectiles. When the enemy ship is at its closest point to the mountain peak (within 2520 m), it can shoot projectiles that can reach the peak.
Considering the height of the mountain peak (1850 m), we can calculate the time it takes for a projectile to travel upwards and hit the peak. We can use the projectile motion equation:
h = v^2 * sin^2(theta) / (2 * g)
where:
- h is the vertical distance traveled
- v is the initial velocity (260 m/s in this case)
- theta is the angle of launch (which we need to find)
- g is the acceleration due to gravity (9.8 m/s^2)
Considering the height of the mountain peak (1850 m) as the vertical distance traveled, we can rearrange the equation to solve for theta:
sin^2(theta) = (2 * g * h) / v^2
sin(theta) = sqrt((2 * g * h) / v^2)
theta = arcsin(sqrt((2 * g * h) / v^2))
Plugging in the values:
g = 9.8 m/s^2
h = 1850 m
v = 260 m/s
theta = arcsin(sqrt((2 * 9.8 * 1850) / (260)^2))
Using a scientific calculator, we find that theta is approximately 26.67 degrees.
Now that we know the angle at which the enemy ship can shoot projectiles to hit the mountain peak, we can determine the horizontal distance traveled by the projectiles.
Using the range equation for the horizontal distance:
R = v^2 * sin(2 * theta) / g
where:
- R is the horizontal distance traveled
- v is the initial velocity (260 m/s in this case)
- theta is the angle of launch (26.67 degrees)
Plugging in the values:
v = 260 m/s
theta = 26.67 degrees
g = 9.8 m/s^2
R = (260)^2 * sin(2 * 26.67) / 9.8
Using a scientific calculator, we find that R is approximately 738.15 m.
Now, let's consider the distance between the mountain peak and the eastern shoreline (305 m). The minimum distance from the eastern shore at which a ship can be safe is equal to the distance traveled by the projectiles (738.15 m) minus the horizontal distance between the mountain peak and the shoreline (305 m):
Minimum distance = 738.15 m - 305 m = 433.15 m
Therefore, the minimum distance from the eastern shore at which a ship can be safe from the bombardment of the enemy ship is approximately 433.15 m.