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Thread: A mass projected vertically upward.

  1. #1

    A mass projected vertically upward.

    A body of mass 7 kg is projected vertically upward with an initial velocity 32 meters per second.

    The gravitational constant is g = 9.8 m/s^2. The air resistance is equal to k*abs(v) where k is a constant.

    Find a formula for the velocity at any time ( in terms of k):

    v(t) = ?

    I'm not really sure how to go about starting this problem. So any help i can get would be appreciated thanks!

  2. #2
    Elite Member
    Join Date
    Jun 2007
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    Re: A mass projected vertically upward.

    Quote Originally Posted by bwat687
    A body of mass 7 kg is projected vertically upward with an initial velocity 32 meters per second.

    The gravitational constant is g = 9.8 m/s^2. The air resistance is equal to k*abs(v) where k is a constant.

    Find a formula for the velocity at any time ( in terms of k):

    v(t) = ?
    going up

    dv/dt = - 9.8 - kv

    going down

    dv/dt = 9.8 - kv

    Now continue...


    I'm not really sure how to go about starting this problem. So any help i can get would be appreciated thanks!
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3

    Re: A mass projected vertically upward.

    Alright so i start with dv/dt = - 9.8(m) - kv m = mass

    From there is this right.

    dv/dt + (1/k)v = -9.8(m)

    Then i found an integrating factor mew(t) = exp of the integral 1/k dt = e^(t/k)

    integral of d/dt(e^(t/k)v) = integral of -9.8(m)e^(t/k) dt

    e^(t/k)v = -9.8k(m)e^(t/k) + c

    then solve for c using the initial condition of v(0) equals 32m/s

    If this isn't right can you give me some direction as to where i went wrong thanks!

  4. #4
    Senior Member
    Join Date
    Nov 2004
    Location
    Seattle
    Posts
    1,397

    Re: A mass projected vertically upward.

    The air resistance is equal to k*abs(v) where k is a constant.
    bwat687,

    Just as an aside, you might want to check your problem statement. Drag is more closely approximated by the square of the velocity. Just a thought...

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