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Physics virgin

  1. Oct 9, 2003 #1
    hey, im new to this forum and physics...i was just wondering if someone could help me with a small question i have on an assignment

    A boy launches an arrow at an initial velocity of 32.5 m/s at an angle of 23.0o above the horizontal.
    How high above the projection point is the arrow after 1.27 s?
  2. jcsd
  3. Oct 9, 2003 #2
    Prijectile Motion

    In this problem you can separate the two dimensions of motion, the X component and the Y component. Neglecting air resistance, the displacement in the X direction is governed by
    X = V * t
    t is time
    V is velocity in the X direction which is constant
    Displacement in the Y direction is governed by
    Y = 1/2 a * t^2 + V * t
    a is acceleration, in this case gravity (-9.8 M/s^2 and negative because the arrow is accelerated downward.)
    t is again time
    V is initial velocity in the Y direction
    Because you know the initial velocity and the angle at which the arrow was fired, you can calculate the initial velocity in the X and Y directions.
    Vx = 32.5 M/s * cos(23.0)
    Vx = 29.9 M/s
    Vy = 32.5 M/s * sin(23.0)
    Vy = 12.7 M/s
    To answer this problem, all you really need in the Y direction.
    Y = 1/2 (-9.8 M/s^2) * (1.27 s)^2 + 12.7 M/s * 1.27 s
    Y = 8.23 M
    So at time = 1.27 sec, the arrow is at a height of 8.23 M.
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