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A $75~\text{kg}$ skydiver can be modelled as a rectangular "box," with dimensions $20~\text{cm} \times 40~\text{cm} \times 180~\text{cm}.$ Given the conditions we may use $F_\text{drag}= {1\over 4}Av^2$. What is his terminal speed if he falls feet first?

I'm not sure what to do with that equation. I found somewhere that at terminal speed the air drag is equal to $mg$, which equal $75~{\text{kg}/\text{m}}$, right?

If that is correct, then could I go about solving the equation in this way?

$$\begin{equation}A=.8m^2\\ F_\text{drag}={.8m^2\over 4}v^2\\ 75~{\text{kg}/\text{m}}={.8m^2\over 4}v^2\end{equation}$$ Then I solved for velocity and got $60.6~{\text{m}/\text{s}}$, but I was looking around on-line and found that a similar problem had a number twice that large, but I didn't recognize any of the formulas they were using, like this one:

$$\text{Drag Force}=.5\rho v^2ACd$$ $Cd$ was the drag coefficient, which I am not provided, and from what I've read it can only be found by observation.

So I'm not sure if I'm headed in the right direction. What am I doing wrong?

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2 Answers 2

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You have at least two errors:

  • The cross-section area of the sky-diver is $$ A=20\,\text{cm} \times 40\,\text{cm} \neq 0.8 \,\text{m}^2 $$
  • The downwards force on the skydiver is $$ F = mg = 75\,\text{kg} \times g \neq 75\,\text{kg/m} $$

Furthermore, when you try to solve problems like this, always work in the (abstract) symbols for things, and proceed as follows:

  • State your approach clearly, e.g. I assume you begin from $$ ma = 0 = F_{\text{gravity}} + F_{\text{drag}} $$ but you don't say. If so, it looks correct, as at terminal velocity the sky-diver is no longer accelerating, $a=0$.
  • Find an expression for what you want to know (in this case $v$) in terms of all the other things ($m$, $A$ etc).
  • Check the dimensions are consistent.
  • Then and only then put the numbers in at the end.

I wouldn't worry about the more complicated equations for the drag force that you found online - the question clearly states the equation you are supposed to use.

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Newton second law can't be treated like that. Remember that this equation has two sides. The LHS tell us about the sum of the forces acting on the body (this is like the cause of the movement), and the RHS tell us about the effect of this forces.

As the body falls, there is only two forces acting on it: its weight $F_w$ and the drag force $F_{drag}$. But the drag force opposes to movement of the body, then, the LHS of Newton second law would be $F_w-F_{drag}$. The RHS would be $ma=m{d^2 x\over dt^2}$.

Then, substituting, you'll find that:

\begin{equation} mg-{1\over 2}A\left({dx\over dt}\right)^2=m{d^2 x\over dt^2} \end{equation}

Now, you need to solve this.

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  • $\begingroup$ isn't this what he tried to do? at terminal, $a=0$. I don't understand what you mean by Newton's law cannot be treated like that. $\endgroup$
    – innisfree
    Commented Oct 16, 2015 at 7:17

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