- #1

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## Main Question or Discussion Point

Hey guys, I am working on deriving some equations of motion for an object with air drag.

I am using this site to help me: http://www.grc.nasa.gov/WWW/K-12/airplane/flteqs.html

Here is my problem, at the bottom of that page they have this equation for horizontal velocity:

1. u = dx/dt = Vt^2 * Uo / (Vt^2 + g * Uo * t)

Then they integrate it to get horizontal distance

2. x = (Vt^2 / g) * ln( (Vt^2 + g * Uo * t) / Vt^2 )

When I take equation 1 and put it in matlab to integrate I get this

3. x = (Vt^2 / g) * ln( (Vt^2 + g * Uo * t) )

Now if I take either equation 3 or 2 and differentiate it, I will get equation 1.

How do they get equation 2 from integrating equation 1? Vt is terminal velocity by the way so that is just a constant. Uo is just the initial velocity. I assume in equation 1 when they took the integral on dx becomes x from x0 to x, x0 being 0 so you just get x. The right hand side you would get from t0 t, t0 being 0 so you are left with t.

Thanks,

Pete

I am using this site to help me: http://www.grc.nasa.gov/WWW/K-12/airplane/flteqs.html

Here is my problem, at the bottom of that page they have this equation for horizontal velocity:

1. u = dx/dt = Vt^2 * Uo / (Vt^2 + g * Uo * t)

Then they integrate it to get horizontal distance

2. x = (Vt^2 / g) * ln( (Vt^2 + g * Uo * t) / Vt^2 )

When I take equation 1 and put it in matlab to integrate I get this

3. x = (Vt^2 / g) * ln( (Vt^2 + g * Uo * t) )

Now if I take either equation 3 or 2 and differentiate it, I will get equation 1.

How do they get equation 2 from integrating equation 1? Vt is terminal velocity by the way so that is just a constant. Uo is just the initial velocity. I assume in equation 1 when they took the integral on dx becomes x from x0 to x, x0 being 0 so you just get x. The right hand side you would get from t0 t, t0 being 0 so you are left with t.

Thanks,

Pete