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Rectilinear Motion/Position Function - Find the distance?

  1. Jun 10, 2007 #1
    Hi! I'm pretty sure I got the first four parts right, but I'm lost on the last one.

    1. The problem statement, all variables and given/known data:

    An arrow is shot from an elevation of 100 meters at a 30-degree angle. It leaves the bow with a speed of 50 meters/second.

    1) Find the initial vertical component of velocity.
    2) Find the vertical position function of the arrow - s(t)
    3) How high will the arrow get?
    4) If the ground is flat, how long will it take for the arrow to hit the ground?
    5) Supposing that horizontal velocity is constant, how far from the shooting point will the arrow be when it hits the ground?

    2. Relevant equations

    Gravity = 9.8 m/s

    a(t) = v'(t)
    v(t) = s'(t)
    s = f(t)

    3. The attempt at a solution

    sin(30) = x/50
    .5 = x/50
    x = 25 meters/second

    v(t)= -9.8t + 50
    s(t)= -4.9t^2 + 50t + 100

    Max height when v = 0
    v(t)= -9.8t + 50
    t = 5.13 seconds
    (5.13) (50 m/s) = 256 m high

    sin(30) = x/256 m
    x = 128m + 100m (<-- initial vertical displacement) = 228 m high

    s(t)= -4.9t^2 + 50t + 100
    Solving for t, the positive constant is is 11.92
    So, it hits the ground after 11.92 seconds.

    I don't know! Help! :)
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 11, 2007 #2


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    Gold Member

    Well you have a function for vertical displacement with respect to time, from question 2 (although i think you might need to have another look at it). You can use this to calculate the time of flight (the time it takes to hit the ground, which according to your equation, you have designated as s=0).

    Once you know how long it is in the air for, you should easily be able to calculate the horizontal displacement, since you are told the initial velocity, and you are also told that horizontal speed is constant.
    Last edited: Jun 11, 2007
  4. Jun 11, 2007 #3
    Right- and I originally just calculated the distance based on the velocity and the time.
    But it's also a factor of the angle it's shot at, so that can't be right. Since it's not going in a strait line, the distance is dependent on the angle it was launched from. (There's an extra credit question which asks to determine the precise angle to achieve the greatest distance, and it was worked out to be around 36 degrees.)
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