Math, Geometry and Physics

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?

The question has two parts. The first is the Maximum downrange distance the enemy ship can shoot. The second is how close can your ship be to western shore, to be safe from the enemy ship (The mountain offers protection).

The Downrange distance is easy.

How do you figure out how close your boat has to be to the western shore, to be protected from enemy fire?

If you get the downrange distance, then subtract (2500 + 300)

You are not understanding the problem.
The peak of the mountain will block many of the higher angle shots taken by the enemy ship.

What formula is available to find the highest angle (We think it is 75 degrees) to clear the peak. Does such a formula exist? Or is this problem trial and error. Remember this is a 10th grade physics question.

OK. Yes, complicated analytically. Trial and error. Iteration is nice on an applet.

http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html
look at the region of 53 degrees.

Thank you - The applet was a great visualization of the motion of a projectile.

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asked by Bob
  1. You use y=xtantheta-((g*x^2)/(2Vo^2cos^2theta))

    Everything is given so you plug it in. 1/cos^2 is 1+tan^2 you get a quadratic from plugging it in and using that trig property. If you solve for it you get 2 angles one is theta low and one is theta high. Id do the math but i don't have a calculator on me

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    posted by Alex
  2. less than 270m or more than 3480m.

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  3. teacher

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    posted by mohammed

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