Please look at question carefully. Max Range is easy. But, there is also a second question. How close to the shore should the boat get, so that it is protected by the peak. What formula can you use to figure that out.
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?
To determine how close a boat needs to be to the western shore to be protected by the mountain peak, we need to consider the trajectory of the projectiles launched by the enemy ship.
Let's assume that the boat is at a horizontal distance x from the western shore. At this distance, the boat is directly across from the peak, which means the line connecting the boat and the peak is perpendicular to the shoreline.
We can start by finding the angle at which the projectiles are launched. To do this, we divide the vertical distance to the peak (1800 meters) by the horizontal distance from the peak to the enemy ship (2500 meters). This gives us an angle of approximately 36.87 degrees (angle A in the diagram below).
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/ |
/ |
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/ A | 1800m
/_____|
2500m
Next, we can split the projectile's initial velocity of 250 m/sec into horizontal and vertical components. The horizontal component (Vx) will be V * cos(A), where V is the initial velocity and A is the launch angle. Similarly, the vertical component (Vy) will be V * sin(A).
Now, let's consider the path of the projectile towards the peak. Since the projectile is launched from the enemy ship, its initial horizontal distance from the peak is equal to the boat's current horizontal distance from the western shore, x.
Using the horizontal component of the velocity (Vx), we can calculate the time it takes for the projectile to reach the peak by dividing the initial distance (x) by the horizontal velocity (Vx).
t = x / Vx
During this time, the projectile will also move vertically. The vertical distance traveled by the projectile is given by the equation:
y = Vy * t - (0.5 * g * t^2)
Where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time calculated above.
Since the vertical distance should be equal to or greater than the height of the peak (1800 meters), we can substitute the values into the equation:
1800 ≤ (V * sin(A) * (x / (V * cos(A)))) - (0.5 * 9.8 * (x / (V * cos(A)))^2)
Simplifying the equation:
1800 ≤ x * tan(A) - (0.5 * 9.8 * (x^2 / (V^2 * cos^2(A))))
Now, we can substitute the given values:
1800 ≤ x * tan(36.87) - (0.5 * 9.8 * (x^2 / (250^2 * cos^2(36.87))))
Solving this equation will give us the possible values for x, which represent the distances from the western shore at which the boat can be safe from bombardment.