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Air drag

  1. Sep 27, 2003 #1
    Hi Everyone! Like always, thank you for helping me figure out my last physics question! I greatly appreciate the help. Here are my next questions:
    Q: Determine the speed at which a baseball must move in order for air resistance to be significant?
    A: mg = ½ * Cd * A * pair * v^2 pair = density of air (1.2kg/m^3)
    m = mass .1488kg
    g = 9.8m/s^2
    A=(pi*r^2)=3.14*.115m^2=.04154 m^2
    Cd = .5 (sphere)

    v= square root of ((2*.1488kg*9.8m/s^2)/(.5* 3.14*.04154 m^2* *1.2kg/m^3 ))
    v = 36 m/s or 80mph

    Q: If the ball were launched vertically into the air at this speed, how high will it go?

    A: After finding the speed in the previous question; I would use this equation to find height:
    KE = PE
    PE = mgy
    KE = 1/2mv^2
    (½)(.1488kg)(36m/s)^2 = (.1488kg)(9.8m/s^2)y
    66.122m=y

    Q: how long will it be air born?

    x = x0 + v0 *t + 1/2at^2
    66.122m = 0m + 36m/s * t + -9.8m/s^2t
    v =

    I am stuck on this part and I don’t know if the rest of it is correct??? Please help!
     
  2. jcsd
  3. Sep 27, 2003 #2

    HallsofIvy

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    Science Advisor

    Actually, I think the first problem is ambiguous. What is meant by "significant". From your equation, you seem to interpret that as meaning the force due to the air resistance is the same as the force due to gravity. If the ball were falling, I could see that: the ball is now no longer accelerating, but going at a constant speed. However, for the ball going up, that makes no sense: gravity and air resistance are both pointing downward.

    Do you have any reason for equating the kinetic energy when the ball is at "terminal speed" downward with potential energy? You do NOT have "conservation of energy" here because there is an outside force (air resistance) acting on the ball.

    What you could do is use force= -mg- ½ * Cd * A * pair * v2 (notice that they are both negative. mdv/dt= force and you have v2on the right hand side so you will have to integrate
    dv/(-mg - ½ * Cd * A * pair * v^2)= dt.
     
  4. Sep 27, 2003 #3
    still confused

    HallsoIvy,

    Thank you so much for your quick response. I hate to bother you, however I am still confused over my last post. My class is an algebra/trig based physics class, therefore, if d stands for the derivative, the equations baffled me. Also, was I on the right track with the first equation? Sorry for my confusion with physics! Thanks again for all your help!
     
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