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Calculating angle of projectile to hit target

  1. Nov 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A projectile is launched from an origin (0cm,0cm) on a tilted plane. The plane makes 30 degrees with the horizontal. The launcher sends the ball with a speed of 0.2m/s.

    At what angle should the projectile be launched to hit the target at (0.7m,0.6m) location?
    Assume there are no resistive forces to the projectile motion in this plane.

    2. Relevant equations
    r_f = r_i + v_i*t + 1/2at2

    Breaking the velocity vector into it's x and y components:
    v_y = vsinΘ
    v_x = vcosΘ

    3. The attempt at a solution
    In the x-direction:
    There is no acceleration in the x direction, so the above equation is reduced to:
    x_f - x_i = vcosΘ*t
    0.6 = vcosΘt
    t = 0.3 / cosΘ (EQUATION 1)

    In the y-direction:
    y_f - y_i = vsinΘt - 4.9t2
    0.7 = vsinΘt - 4.9t2 (EQUATION 2)

    SUB EQUATION 1 INTO 2

    0.7 = 0.2sinΘ (0.3/cosΘ) - 4.9(0.3/cosΘ)2
    This is where I get stuck... solving the trigonometric equation. I can't find a solution, so I'm thinking that my above reasoning may be wrong.
    Either that, or I am just having an incredibly hard time with trig -- which I've been out of practice for 8 years.
    Any suggestions / guidance is very much appreciated.
     
  2. jcsd
  3. Nov 18, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    From the description in the problem statement it sounds like the projectile is being launched in the tilted plane. If that's so you'll want to consider what the acceleration due to gravity is in the plane. It won't be 9.8 m/s2, but rather some fraction of that which depends upon the tilt of the plane to the horizontal.

    For your equation 1 it looks like you've divided 0.6 by 0.2 and arrived at 0.3 (the x-displacement 0.6 m divided by the initial velocity of 0.2 m/s). You'll want to check that.

    These trig equations can be tricky. Keep in mind the basic identity sin2 + cos2 = 1. If you can get everything in terms of either sin or cos then it'll boil down to algebra if you make suitable substitutions (for example: "let z = cos2(Θ)").
     
  4. Nov 19, 2015 #3
    Thank you, this helped a lot! Expressing the trig in terms of another variable was the trick at the end.
     
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