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Launch Angles

  1. Feb 3, 2006 #1
    Can anyone help with this problem??
    Children bring their old pumpkins to a tower and compete for an accuracy in hitting a target on the ground. Suppose the tower height is h=8.0m and the bull's eye on the ground is a distance d=3.4 m from the launch point. (neglect air resistance.)

    If the pumpkin is given an initial horizontal speed of 3.3 m/s, what are the direction and magnitude of its velocity at the following moments?

    a.) .75s after launch (find direction of magnitude and velocity.)

    b.) just before it lands (find direction of magnitude and velocity.)

  2. jcsd
  3. Feb 3, 2006 #2
    Use [tex]v = v_0t + \frac{1}{2}at^2[/tex] to find the vertical component of velocity (v_0 is the initial velocity, which is zero in the vertical plane, and a = g), and then to find the magnitude use [tex]a^2 = b^2 + c^2[/tex] (the other part is the horizontal component - which is constant). The direction is then simply the inverse tan of the two components.
  4. Feb 3, 2006 #3
    Sorry, that equation I gave you doesn't make sense.
    [tex] v = v_0t + at[/tex]
    The displacement equation is:
    [tex] s = v_0t + \frac{1}{2}at^2 (+ h_0)[/tex]
    (when a is constant)
    v means velocity, v_0 means initial velocity, s means displacement, a means acceleration, and t means the time. h_0 is the initial displacement from the origin. In terms of the vertical component, this will be 8. This will not apply for the horizontal component.

    for #2, solve the equation to get t, and then use the method I stated earlier.

    Don't know why I tried to combine the two, I just wasn't concentrating.
    Last edited by a moderator: Feb 3, 2006
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