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Projectile motion with air resistance

  1. Sep 27, 2005 #1
    Ok, in this problem, only the y direction is dealt with

    Here is the problem

    You are standing on top of a building, 30 m above the ground. You throw a ball (m = 0.15 kg) with an initial velocity or 20 m/s (in the y direction). Air resistance is given as (1/30)*v (yet again only considering the y direction because range is not an issue here).

    a) find the maximum height of the ball
    b) find the time t when the ball hits the ground 30 m below you

    Ok so I set it up like this

    mv' = -mg - v(1/30)
    v' = -g - v(1/30m)

    from here I get kind of confused.

    I tried this

    v' + v(1/30m) = -g
    (e^(t/30m)v)' = -ge^(t/30m)
    integrating:
    e^(t/30m)v = -30mge^(t/30m) + c
    v = -30mg + c/(e^(t/30m))

    I'm pretty sure I'm wrong by here. Help would be appreciated.
     
  2. jcsd
  3. Sep 28, 2005 #2

    Tide

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    That will give you the time it takes to reach the top of the trajectory. Don't forget that the direction of the drag force reverses on the way down so you'll have a different equation to solve for that part.
     
  4. Sep 28, 2005 #3
    well..I don't even think I did it right

    When I try to solve for t for the intial codition of V(0) = 20, and then set V = 0 in order to solve for the time at which the ball has reached its peak, I get a negative time. So I assume my equation is wrong..dunno what to do though
     
    Last edited: Sep 28, 2005
  5. Sep 28, 2005 #4

    Tide

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    You didn't apply your initial condition properly when you integrated.
     
  6. Sep 28, 2005 #5
    I tried everything I could, including

    v' = -g - kv where k = (1/30m)
    v'/(-g-kv) = 1
    v'/(g/k + v) = -k
    integrating with following limits
    ln(g/k + v)|from v_o to v = -kt (from 0 to t)
    ln(g/k + v) - ln(g/k + v_o) = -kt
    ln [(g/k + v)/(g/k+v_o)] = -kt
    g/k + v = e^(-kt)(g/k + v_o)
    v = e^(-kt)(g/k+v_o) - g/k
    Is it right
    If it is, then integrating again would this also be right
    y-y_o = -(1/k)e^(-kt)(g/k+v_o) - (g/k)t
    y = -(1/k)e^(-kt)(g/k+v_o) - (g/k)t + y_o

    I think this 'looks' right but when I try to calculate the displacement for the time I calculated where the ball should have a velocity of 0 (about 1.6835 s), by plugging into the second equation, I get a very negative value..Thanks in advance
     
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