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    Problems with integration word problems

    I was absent the day that we covered these, and have yet to have received a sufficient explanation of how to do them. I understand integration fairly well, but I'm unsure of how to exactly apply it in these problems. One excellent example is this:

    The Washington Monument is around 550 ft. tall. If you're trying to throw an object from gound level to the top of the Monument, what would your initial velocity need to be? Gravity is at -32 ft/s, don't worry about air resistance.

    I understand how to integrate, but I don't know how to begin to apply it to this kind of a problem. Thanks in advance for all the advice.
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    Quote Originally Posted by bobsanchez View Post
    I was absent the day that we covered these, and have yet to have received a sufficient explanation of how to do them. I understand integration fairly well, but I'm unsure of how to exactly apply it in these problems. One excellent example is this:

    The Washington Monument is around 550 ft. tall. If you're trying to throw an object from gound level to the top of the Monument, what would your initial velocity need to be? Gravity is at -32 ft/s, don't worry about air resistance.

    I understand how to integrate, but I don't know how to begin to apply it to this kind of a problem. Thanks in advance for all the advice.
    a = -32

    antiderivative of acceleration is velocity ...

    v = -32t + C

    at t = 0 , velocity = v_0 , initial velocity.

    v = v_0 - 32t

    antiderivative of velocity is position ...

    y = v_0 t - 16t^2 + C

    if the starting position is the ground, C = 0

    y = v_0 t - 16t^2

    height of the monument is 550 ft ...

    550 = v_0 t - 16t^2

    at the top of its trajectory, v = 0 ... t = \frac{v_0}{32}

    solve for v_0
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    So you basically just integrate acceleration, then subtract that from what your velocity, obtain the position, and solve for whatever it is that you're looking for? Okay, that makes sense.
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    Quote Originally Posted by bobsanchez View Post
    So you basically just integrate acceleration, then subtract that from what your velocity, obtain the position, and solve for whatever it is that you're looking for? Okay, that makes sense.
    huh?
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    Quote Originally Posted by skeeter View Post
    huh?
    Oops. Initial velocity, rather.
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    Quote Originally Posted by bobsanchez View Post
    Oops. Initial velocity, rather.
    no "subtraction" ... the negative sign is there to indicate the direction of the acceleration (down).

    understand that v_0 is the constant of integration for the antiderivative of a = -32.
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