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Physics projectile angle Question

  1. Mar 23, 2003 #1
    An enemy ship is on the western side of a mountain island. The enemy ship can maneuver to within 2500 m of the 1800-m-high mountain peak and can shoot projectiles with an initial speed of 250m/s. If the eastern shoreline is horizontally 300m from the peak, what are the distances from the eastern shore at which a ship can be safe from the bombardment of the enemy ship?

    I need to find the two angles at which the height of the projectile is 1800m at 2500m from the boad, but I'm having trouble relating all the different equations for Range, max height, and the other kinematic equations for motion in two dimensions.

    A suggestions would be greatly appreciated.
  2. jcsd
  3. Mar 23, 2003 #2
    OK, here's my latest approach. I want the height of the projectile to be 1800m at the same time it's moved 2500m horizontally.

    The time at which it moves 2500m horizontally is 10sec/cos@. I then plug this into y = at^2 + vt + y(initial).

    So, 1800m = -490m/(cos@)^2 + 2500m(sin@)/cos@ Am I correct so far?

    If so, I'm at a loss as to how I should deal with the trigonometry.
  4. Mar 24, 2003 #3
    Hi discoverer02,
    you forgot the factor 1/2 in your equation, but you fixed that in the next step. So I think your analysis is OK.

    I tried more simple approaches, but none worked. So I think this is the way to do it. Next, you could of course substitute sin a = sqrt (1 - (cos a)^2). This will give a biquadratic equation for cos a, which has indeed 2 solutions. It's not difficult, but lengthy & boring. Sorry, I see no other way.
  5. Mar 24, 2003 #4
    Hi arcnets,

    Thanks for taking the time to look at the problem and posting a suggestion.

    I'm still a little stumped by the trigonometry. If I use the identity you suggest, do I plug sqrt(1-(cos A)^2) into the quadratic equation when solving for 1/(cos a)? If not, then I can't figure out how to get rid of it.

    Thanks again for your help.

  6. Mar 24, 2003 #5
    Next (omitting units and using C = cos@):
    1800 = -490/C^2 + 2500 sqrt(1-C^2)/C
    1800 C^2 = -490 + 2500 C sqrt(1-C^2)
    1800 C^2 + 490 = 2500 C sqrt(1-C^2)
    1800^2 C^4 + 490^2 + 2*1800*490*C^2 = 2500^2 C^2 (1 - C^2)
    1800^2 C^4 + 490^2 + 2*1800*490*C^2 = 2500^2 C^2 - 2500^2 C^4
    4300 C^4 + (2*1800*490 - 2500^2) C^2 + 490^2 = 0.

    That's the biquadratic beast. Solve.
  7. Mar 24, 2003 #6

    Thanks again.

    I thought maybe there was an easier way than squaring both sides that I wasn't seeing, but I guess not.

    I guess I've become use to homework problems resolving themselves into neat little equations. Not always the case though.
  8. Mar 25, 2003 #7
    Maybe there's a nice, easy, and symmetric solution. But I don't see one.
    There's an error in my last line, Should read
    9490000 C^4 + ...
  9. Mar 26, 2003 #8

    use y=y1+V1y(t-t1)-g/2(t-t1)^2 (assume y=0 and t1=0)
    which gives V1y+-g/2(t^2)

    x=x1+V1x(t-t1) (assume X1=0 and t1=0
    which gives V1x(t)

    thus the entire formula to determin the angle(upper) and angle(lower) is:
    y=V1y(x/V1*x) - g/2(x^2/V1x^2)

    for y you should work everything down to a quadratic equation.

    Get there using the formula:

    (Y1 sin(theta1))/Y1 cos(theta1))*x - g/2V1^2(x^2/cos^2(theta1)

    You can see that sin and cos turn into xtan(theta1)

    you can pull out the x^2 on the top of the eqation on the right side [g/2(x^2/V1x^2)] and get 1/cos^2(theta1) This will give you (being that 1 is the same as guess what? cos^2(theta1)+sin^2(theta1) just plug in that in to your formula and that will eventually give you the ultimate quadratic of:

    which turns into the formula I have post on my web site.

    http://home.satx.rr.com/wysocki/images/formula.jpg [Broken]

    That should get you going. Just remember that the + value of the quadratic is the upper angle and the - value is the lower angle.

    Im not going to give you exact values but the upper angle is more than 69 degrees and the lower angle is less than 50 degrees.

    Peace out bro

    Last edited by a moderator: May 1, 2017
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