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Math Help - Quick question about position due to gravity.

  1. #1
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    Quick question about position due to gravity.

    I have a problem where I'm given a bunch of info. about a ballon dropping a bomb, etc. The thing is that I am not given any functions but I know that I have to use the function for position of falling bodies with respect to time.
    It's been a while since I saw this formula, so i'm not sure if this is right:

    In feet: s(t)=-16t^2 + v_0 t + s_0
    where v_0 is initial velocity and s_0 is starting position.

    Is any of this right?
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  2. #2
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    I'm not sure what you are trying to do.

    You have a bomb floating to the earth on a balloon, and you have to construct the equation which describes the displacement the balloon as fallen from its original position after time t?
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  3. #3
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    Quote Originally Posted by Arturo_026 View Post
    I have a problem where I'm given a bunch of info. about a ballon dropping a bomb, etc. The thing is that I am not given any functions but I know that I have to use the function for position of falling bodies with respect to time.
    It's been a while since I saw this formula, so i'm not sure if this is right:

    In feet: s(t)=-16t^2 + v_0 t + s_0
    where v_0 is initial velocity and s_0 is starting position.

    Is any of this right?
    None of what you're asking can be answered without knowing the question. It's always better to post the original question instead of hoping that our crystal balls are in good working order.
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  4. #4
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    Try one (or more) of the kinematic equations:


    v = v_0 + at

    y - y_0 = v_0t + \frac{1}{2}t^2

    v^2 = v_0^2 + 2a(y - y_0)

    y - y_0 = \frac{1}{2}(v_0 + v)t


    Where v is the final velocity, v_0 is the initial velocity, y is the final height, y_0 is the initial height, a is the acceleration due to gravity near Earth's surface ( g = 9.8 \frac{m}{s^2}), and t is the time.

    Patrick
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    None of what you're asking can be answered without knowing the question. It's always better to post the original question instead of hoping that our crystal balls are in good working order.
    Ok, here is the problem:

    A bomb is dropped from a ballon hovering at an altitude of 800 ft. From directly below the balloon, a projectile is fired directly upward toward the bomb exactly 2 seconds after the bomb is released. With what initial speed should the projectile be fired in order to hit the bomb at an altitude of exactly 400 ft.?
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  6. #6
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    The problem is for some reason in the Standard system (feet) instead of the Metric (meters), so be sure to consider that when solving the problem (This is where the -16 comes from in your original equation).

    The bomb accelerates downward at a rate of 9.8 \frac{m}{s^2} or just over 32 \frac{ft}{s^2}. After t seconds, it will be halfway down, from 800 ft to 400ft. At this precise moment, the projectile from the surface will intercept it. Therefore, you need to solve for t, the time it takes the bomb to drop 400 ft, from an initial velocity of zero.

    Then, when you have t, you must subtract 2 seconds to get the time for the projectile's flight from the surface to the bomb, over a distance of 400 ft. When you have this time, you can then use the kinematic equations to solve for the initial velocity, knowing that the acceleration due to gravity will slow the projectile so that the final velocity is less than the initial velocity.

    You will have the variables of t, a, and y. Using the second equation listed (a is not needed), you can solve for the initial velocity. Let me know what the answer is.

    Patrick
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  7. #7
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    I get t=5 for the bomb to be at an altitude of 400 ft. and when t=3 the projectile needs to be launched with an initial velocity of 181.33 ft. per second.
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