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Math Help - Falling object

  1. #1
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    Falling object

    A coin is dropped from a height of 750 feet.The height, s, (measured in feet), at time, t (measured in seconds), is given by s = - 16t^2 +750

    • Find the average velocity on the interval [1,3]
    • Find the instantaneous velocity when t = 3
    • How long does it take for the coin to hit the ground?
    • Find the velocity of the coin when it hits the ground.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Samantha View Post
    A coin is dropped from a height of 750 feet.The height, s, (measured in feet), at time, t (measured in seconds), is given by s = - 16t^2 +750

    You are expected to know that:

    s(t) = position function
    v(t) = s'(t) = velocity function
    a(t) = v'(t) = acceleration function

    Given: s(t) = -16t^2 + 750

    \Rightarrow v(t) = -32t

    \Rightarrow a(t) = -32
    • Find the average velocity on the interval [1,3]
    The average velocity is given by the slope of the secant line connecting the two points.

    Thus, \mbox { Average Velocity } = \frac {s(3) - s(1)}{3 - 1}

    Can you take it from there?

    • Find the instantaneous velocity when t = 3
    We simply evaluate v(3), that is, the velocity function when t = 3. I found that for you above

    Can you take it from there?


    • How long does it take for the coin to hit the ground?
    s(t) gives the position function. we want the time when the position is zero, therefore, we set s(t) = 0 and solve for t, this will give you the required time. To be more explicit, we want t such that:

    -16t^2 + 750 = 0

    Can you take it from here?


    • Find the velocity of the coin when it hits the ground.
    For this question, we simply plug in the value of t we found in the previous question into the velocity function

    Can you take it from here?
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  3. #3
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    I'm very bad in it...

    Can you plz finish the problem.

    I would really appreciate...thank you very much
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  4. #4
    Bar0n janvdl's Avatar
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    How long does it take for the coin to hit the ground?
     s = -16t^2 + 750

    Where s = 0 ; that is where the ground is.

     0 = -16t^2 + 750

     16t^2 = 750

    Thus  t = +6,85 or  t = -6,85

    But  t = -6,85 cannot be true.

    Therefore only  t = +6,85 is true.
    Last edited by janvdl; July 2nd 2007 at 01:32 PM.
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  5. #5
    Bar0n janvdl's Avatar
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    Find the velocity of the coin when it hits the ground.
    We know it hits the ground in 46,875 seconds. We calculated that above.

    But now we need to differentiate  s

    Remember the following:
    When S(distance) is differentiated, we get V(velocity).
    When V is differentiated, we get A(acceleration).

    ----------

    So:

     \frac{dS}{dT} = V = -32t

    We know t = 6,85 seconds.

    So  V = -32(6,85)

     V = -219,2 \ feet \ per \ second

    What does the negative mean? It merely means the coin is moving downwards.
    Last edited by janvdl; July 2nd 2007 at 01:33 PM.
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