1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Projectile motion and angle

  1. Sep 9, 2003 #1
    I need help putting this problem into a workable equation:

    A projectile is launched from ground level at an angle of 12 degrees above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, w/o changing the launch speed, so that the range doubles?

    So far I tried this:

    I broke the problem into its components: x & y.
    FOR X: {x=Vo*COS(12)t Vox=Vo*Cos(12) a=0 t=t}
    FOR Y: {y=? Voy=Vo*SIN(12) a=-9.8 m/s^2 t=t}

    Then I used t=x/Vo*COS(12), then substituted that for T in the Y parts, so that:

    y= tan(12)x - [(4.9 x^2)/(Vo^2*cos(12)^2)]

    but then I got: x=tan(12)y, which doesn't help me at all.

    I'm guessing I'm approaching this problem in a totally WRONG way!
  2. jcsd
  3. Sep 9, 2003 #2


    User Avatar
    Staff Emeritus

    so you start out with a vector v-> and an angle theta1.

    x = |v->| * cos(theta1) * t
    y = |v->| * sin(theta1) * t - (1/2) * g * t^2

    when y = 0, the projectile will be on the ground

    0 = |v->| * sine(theta1) * t - (1/2) * g * t^2

    t = 0 , 2 *|v->| * sin(theta1) / g

    when t = 0 is at the start, so the other solution is the time it hits the ground

    x = |v->| * cos(theta1) * 2 *|v->| * sin(theta1) / g

    pluging in t into the x equation and doing some math you get:

    x = |v->|^2 * sin( 2 * theta1) / g

    If you want the range to be double then it is 2 times the x equation with the new angle.

    x2 = 2 * |v->|^2 * sin( 2 * theta2) / g

    equate x and x2 to find theta2. You know theta1 = 12 degrees

    x = x2 , then theta2 = 5.87 degrees

    And there is your answer
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook