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  1. #1
    AAA
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    Question falling object with air resistance

    (eight-pound object dropped from a height of 5000 feet, where air resistance is proportional to the velocity.)

    write the velocity as a function of time if its velocity after 5 secs is approximately -101 feet per second.
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    Quote Originally Posted by AAA View Post
    (eight-pound object dropped from a height of 5000 feet, where air resistance is proportional to the velocity.)

    write the velocity as a function of time if its velocity after 5 secs is approximately -101 feet per second.
    F= ma so (8/32)a= (1/4)dv/dt= -32+ kv. v(0)= 0. Can you solve that equation for v? It will contain a k of course so you need to use v(5)= -101 to find k.
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    Quote Originally Posted by HallsofIvy View Post
    F= ma so (8/32)a= (1/4)dv/dt= -32+ kv. v(0)= 0. Can you solve that equation for v? It will contain a k of course so you need to use v(5)= -101 to find k.

    Thanks, but why 8/32?
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    F = mg

    8 lbs is the Force due to gravity.

    8 = m(32) ... acceleration due to gravity is 32 ft/s^2

    m = 8/32 = 1/4 slug

    yes, slugs ... one of the joys of the English system of measurement
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    Object falling with air resistance

    Quote Originally Posted by skeeter View Post
    ... yes, slugs ... one of the joys of the English system of measurement
    True, but don't blame us - we use SI units, and have done for 40 years or so! (All right, they are a bit French, but we all have to swallow our pride at some time or other!)

    In fact, slugs are horrible creatures, best avoided. Just call the mass m - it cancels anyway.

    The equation of motion is then:

    -mg + k'v = m\frac{dv}{dt}, where v is measured vertically upwards

    And you can then divide through by m, and replace \frac{k'}{m} by k to obtain:

    -g + kv = \frac{dv}{dt}

    After separating the variables, integrating, and using v(0) = 0, this gives:

    v = \frac{g}{k}(1 - e^{kt})

    So when t = 5, v=-101 (and g = 32):

    -101 = \frac{32}{k}(1 - e^{5k})

    How you are expected to solve this equation for k, I don't know (Newton's method?). In fact, I found k \approx -0.2 (but I cheated and used a spreadsheet!)

    So the solution is

    v\approx 160(e^{-0.2t}-1)

    Grandad
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