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Projectile motion, vector components

  1. Jul 7, 2010 #1
    1. The problem statement, all variables and given/known data

    A child throws a snowball with a horizontal velocity of 18m/s directly toward a tree, from a distance of 9.0m and a height above the ground of 1.5m.

    a) After what time interval does the snowball hit the tree?
    b) At what height above the ground does the snowball hit the tree?

    2. Relevant equations

    displacement on x = horizontal velocity * time

    displacement on y = initial vertical velocity * time + 0.5 * acceleration * time squared

    or

    displacement on y = final vertical velocity * time - 0.5 * acceleration * time squared

    3. The attempt at a solution

    For a), using the first formula, the calculation is:
    9.0 m = 18m/s * t
    t= 0.5s

    Now I'm not sure what to do with b). If I use the equation:

    displacement on y = initial vertical velocity * time + 0.5 * acceleration * time squared

    is the initial vertical velocity 0? Since the boy throws the snowball horizontally? I think it is, but I don't get the right answer when I do this (the answer is 0.3 m).

    What I'm doing is:

    displacement on y = (0) * (0.5) + 0.5 * (-9.8) * (0.5)^2
     
  2. jcsd
  3. Jul 7, 2010 #2
    "a" answer is correct

    for "b" you can use:
    [tex]Y=Y_{0}+V_{0}t+\frac{1}{2}at^{2}[/tex]
    where Y is the final altitude, [tex]Y_{0}[/tex] equals the initial altitude (1.5m), a the acceleration (a=g=-9,8(m/s^2)) and t is the time.

    you resolve that and you'll get Y=0,275 (m)
     
  4. Jul 7, 2010 #3

    Doc Al

    User Avatar

    Staff: Mentor

    That will give you the displacement from the original height of the snowball when it was thrown. How high was that above the ground?
     
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