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Air resistance (pebble throwing)

  1. Mar 26, 2007 #1
    i have solved the question of a pebble thrown upwards without air resistance
    speed 10m/s mass 0.02kg ma=w, w=-mg

    man on cliff 50 feet high throws pebble upwards so it finishes in sea,
    what height does it reach v=0 t = 10/g approx =1
    since motion started at t =0 & duration of time 1s
    x = 100/g - 100/2g is approx 5.1m so maximum height 55.1 m

    BUT
    i now need to find height with air resistance 0.008 v
    any pointers to x value warmly recieved
    cheersjk
     
  2. jcsd
  3. Mar 26, 2007 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Again, mass times acceleration equals force.

    Without air resistance, the only force its weight: -mg so you have
    [tex]m\frac{dv}{dt}= -mg[/tex]
    The easy solution to that is v= -gt+ C. Since the initial velocity is 10 m/s
    v(t)= dx/dt= -gt+ 10. Integrate again to get x as a function of t. Solve v(t)= -gt+ 10 to find when it stops going up, put that value of t into the x fuction to find the highest point.

    I said that to review what you must have done. Now, with air resistance, the force is -mg- 0.008v (negative because friction force always opposes v).

    Your equation is
    [tex]m\frac{dv}{dt}= -mg- 0.008v[/tex]
    You can rewrite that as
    [tex]\frac{mdv}{mg+0.008v}= -dt[/itex]
    and integrate both sides to find v as a function of t. From there the rest of the problem is the same.
     
  4. Mar 26, 2007 #3
    i understand your reasoning, thank you

    cheers jk
     
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