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Math Help - Finding the difference in range of a projectile (air resistance vs. free)

  1. #1
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    Finding the difference in range of a projectile (air resistance vs. free)

    We were give two equations that show the position as a function of time on the x and y axis where there is a linear drag force of 'beta*a vector v'. I have already solved for the case of no air resistance. The following are the two equations which include air resistance:
    x(t) = x(t=0) + ((m*vo(x))/beta)*(1-e^((-beta/m)*t))
    --> x(t=0) is 0, giving:

    x(t) = [(m*vo(x))/beta]*[1-e^((-beta/m)*t)]

    y(t) = [(vo(y)*(m/beta))+g*((m/beta)^2)]*[1-e^((-beta/m)*t)]-(g*(m/beta)*t)
    --> y(t) would be 0, considering the full path of the trajectory results in no delta y, so setting the equation equal to 0, I was able to start solving for 't' to plug into the equation for x(t) to find a total Range.

    However, I am caught on the exponential function since it includes 't'. I am not sure how to get that out of there so I can solve for 't' without including 't' in the result. This is where I am so far:

    t = [(vo(y)+m)/g*beta]*[beta*(1-e^((-beta/m)*t))]

    If you all could help me, I would greatly appreciate it! I'm sorry if there is a way to input equations, I am new to the forum and don't know how to do that if there is a way.

    Thanks everyone!
    -Tina
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Finding the difference in range of a projectile (air resistance vs. free)

    You don't need to eliminate t in order to find a single function of y=f(x), but rather you have x(t) and y(t) as two parametric functions which define your projectile and you can plot it too.

    It means you vary t and calculate x(t) & y(t) that give your projectile.

    Assuming:
    m=10
    v0x=1
    v0y=10
    beta=3
    g=9.8

    Trajectory would look like:

    Attached Thumbnails Attached Thumbnails Finding the difference in range of a projectile (air resistance vs. free)-projectile.png  
    Last edited by MaxJasper; August 23rd 2012 at 11:28 PM.
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