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Homework Help: Projectile Motion of trebuchet

  1. May 26, 2007 #1
    1. The problem statement, all variables and given/known data

    I'm looking at trebuchets. Basically, a trebuchet shoots a projectile with an initial velocity of 30ms-1 at an angle of 55, 65 and 75 degrees. Friction/Wind resistance do not account.

    I need to find the maximum height reached, the time the projectile spends in the air and the range. I am having troubles with the time spent in air, therefore affecting my last calculation.

    I got a maximum height of 30.811m at 55 degrees.
    37.71m at 65 degrees.
    42.842m at 75 degrees.

    2. Relevant equations

    As mentioned, I'm having trouble finding the total time spent in air. I know I have he following variables. I know that the time to reach max height x2 is the time of flight.

    Y = Verticle, X = Horizontal

    Uy= 30ms-1
    Vy = 0ms-1 (Obviously no movement at the exact highest point)
    a= -9.8ms-2
    Sy = 30.811m

    I tried using the rule V = u+at


    This gives me (0-30)/-9.8
    Giving me 3.06 seconds and a total time of 6.122 seconds.

    3. The attempt at a solution

    If I wanted to use that equation to find out the time taken if launched at 65 and 75 degrees, I would get the same answer due to the fact that verticle displacement isnt accounted.

    I've been going over this again and again and I can't figure out anything.
    Last edited: May 26, 2007
  2. jcsd
  3. May 27, 2007 #2
    Well actually there is movement at the highest point.
    It has vertical and horizontal speed. The vertical speed is zero at the highest point, but the horizontal speed is constant.
    You said that "Y = Verticle, X = Horizontal", but you never seem to notice the horizontal.
    When the angle is 55 for example, it has a vertical and a horizontal component.
    Vy = sin 55 * v
    Vx = cos 55 * v
    So Uy is not 30ms-1.
    I think you can solve it now.

    EDIT: I just noticed that the values for maximum height are correct. So that's what you did (used vertical and horizontal components).
    But then why are you using 30 m/s for (0-30)/-9.8 ? Shouldn't it still be sin 55 * 30?
    You are only doing calculations vertically.
    The acceleration only acts vertically, the final speed is zero only vertically, so then why use the resultant speed in the formula. You have to use the initial vertical speed as well.
    Last edited: May 27, 2007
  4. May 27, 2007 #3
    Thanks, I just realised I wasn't calculating both horizontal and vertical attributes. Can't beleive I missed that!

    Thanks again :D
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