1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding launch angle from vertical and horizontal distance

  1. Oct 31, 2015 #1
    1. The problem statement, all variables and given/known data
    The mountain rescue officers trigger the avalanche by firing a gun at a distant mountain slope.
    The muzzle velcoity of the gun is 225m/s. The horizontal and vertical distance from the gun to the slope is
    x=1,072m and y=538m respectively. Find the lowest angle between the barrel of the gun and the horizontal plane.

    2. Relevant equations
    rf = ri + vt + 1/2at2
    Dividing each into the horizontal and vertical components:
    - Horizontal component: xf = xi + vxt + 1/2axt2
    - Vertical component: yf = yi + vyt + 1/2ayt2

    Vxi = VicosΘ

    Vyi = VisinΘ

    3. The attempt at a solution
    I started off by manipulating the equations in terms of time (t), since I don't have that either.

    In the x direction, the bullet is under constant velocity, so a=0; xf = 1072m; xi = 0
    xf = xi + vxt + 1/2axt2
    1072 = 225cosΘ t
    (1) t = 1072 / 225cosΘ

    In the y direction, the bullet is under constant acceleration due to gravity, ay = -g = -9.8m/s2
    yf = yi + vyt + 1/2ayt2
    528 = 225sinΘ t - 4.9t2
    (2) 538 = t(225sinΘ - 4.9t)

    Then I substituted equation (1) into (2)...
    538 = [1072/225cosΘ] [225sinΘ - 4.9(1072/225cosΘ)
    538 = 1072sinΘ/cosΘ - 111.22966/cos2Θ
    and this is where the trig substitutions start getting messy. It gets quite long, and I ended up with 1.211249 = sin2Θ, which is impossible.

    So either I am approaching the question wrong, or I am not manipulating my trig properly.

    Any help or direction would be greatly appreciated.

     
  2. jcsd
  3. Oct 31, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    That has an actual solution, so your mistake is somewhere in the part you didn't show.

    Solving for the cosine is probably a bit easier.
     
  4. Oct 31, 2015 #3

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    It's not clear how you went from this:
    538 = 1072sinΘ/cosΘ - 111.22966/cos2Θ

    to this:
    1.211249 = sin2Θ

    It would seem that there is a problem with your algebra of trig here which need double checking.
     
  5. Oct 31, 2015 #4
    If it helps any the equation for the distance along the plane is
    D = 2 * v^2 ^ cos^2 alpha * (tan alpha - tan beta) / (g * cos beta)
    where alpha is the angle of the projectile and beta the angle of the plane.
    It would seem that the solution for this particular configuration could be found without deriving this formula,
    but I admit to never having tried it.
    Incidentally, the equation is derived using
    y = - g t^2 /2 + v t sin alpha
    x = v t cos alpha
    and eliminating the time from these equations.
    Good Luck and Happy Halloween!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted