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Projectile motion problem

  1. Jul 1, 2013 #1
    A cannon with a muzzle speed of 1000 m/s is used to start an avalanche on a mountain slope. The target is 2000 m from the cannon horizontally and 800 m above the cannon. At what angle, above the horizontal, should the cannon be fired?
  2. jcsd
  3. Jul 1, 2013 #2
    Where is your attempt at a solution. No try, no help.
  4. Jul 1, 2013 #3
    lol, ok.

    help me solve this then :


    ^ le attempt xD
  5. Jul 1, 2013 #4
    I believe this equation is correct. To solve, you must use a trig identity to arrive at a quadratic equation to solve.
  6. Jul 1, 2013 #5
    how do I do that? xD

    the reason why I got stuck is because of that junk.

    I need help solving that or an alternative method to solve for the angle.
  7. Jul 1, 2013 #6
    Well, if you don't want to do the trig thing, then use a graphing calculator.
  8. Jul 1, 2013 #7
    we( the class) cant "graphing calculator" O:
  9. Jul 1, 2013 #8


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    It looks almost right. Please double-check.
  10. Jul 1, 2013 #9
    This is correct. OK, I'll give another hint. Divide everything by cos^2(x) then look for the trig identity to make a quadratic equation to solve.
  11. Jul 1, 2013 #10
    looks very right ^.^
  12. Jul 1, 2013 #11


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    I'm having trouble understanding the origin of the 1000 factor:


    I'll check my math again in the morning--with fresher eyes.

    barryj's suggestion is a good one.
  13. Jul 2, 2013 #12


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    yeah, I don't think that 1000 should be there either
  14. Jul 2, 2013 #13
    The 1000 should not be there. Did you all figure out the required trig idenity?
  15. Jul 2, 2013 #14
    There is something wrong with the equation you gave as others says.
    Btw does it says you to use 9.8 for g? If not use 10 instead of 9.8
  16. Jul 2, 2013 #15
    I was taught this weeks ago in my college.
    Teacher got the answer by derivation.
    Last answer was tan@ = u / (u^2 - 2gh)^1/2
    Where u is 1000
    g is gravity
    h is 800
  17. Jul 2, 2013 #16
    Consider that sec^2(x) = 1 + tan^2(x)
  18. Jul 2, 2013 #17
    wrong answer ^.^
  19. Jul 2, 2013 #18
    Post 16 is the clue!
  20. Jul 2, 2013 #19


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    Unfortunately, this answer does not contain a variable for horizontal distance (2000m). So...it's not helping. Even if you provided the correct answer, it probably would not help the OP who is currently trying to understand how to determine the answer, with the help of the powerful clues provided by barryj.
  21. Jul 2, 2013 #20


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