Opening a parachute

To establish an equation for the frictional force, we use the dimensional analysis. The relevant quantities each have the units $\begin{aligned} [F] &= \text{N} = \text{kg} \cdot \text{m} \cdot \text{s}^{-2}\\ [\rho] &= \text{kg} \cdot \text{m}^{-3} \\ [A] &= \text{m}^2 \\ [v] &= \text{m} \cdot \text{s}^{-1} \\ \end{aligned}$ Now we try to create a force from the quantities $\rho$ , $A$ and $v$ : $\begin{aligned} & & [F] &\stackrel{!}{=} [\rho]^\alpha \cdot [A]^\beta \cdot [v]^\gamma \\ \Rightarrow & & \text{kg} \cdot \text{m} \cdot \text{s}^{-2} &= \text{kg}^\alpha \cdot \text{m}^{-3\alpha + 2 \beta + \gamma} \cdot \text{s}^{-\gamma} \\ \Rightarrow & & (\alpha,\beta,\gamma) &= (1,1,2) \\ \Rightarrow & & F_f &= c \cdot \rho \cdot A \cdot v^2, \quad [c] = 1 \end{aligned}$ The constant $c$ is dimensionless. This is the only possible form that the friction force equation can have, because only in this way can we get a quantity with a unit of force. However, this does not apply if the viscosity of the air also plays a role, because then an additional parameter has to be considered, especially in the limit of high viscosity, so that in this case applies $F_f \propto v$ .

After a long fall the parachutist reaches a final velocity $v_ \infty = \lim_ {t \to \infty} v (t)$ , so that the sum of gravitational force and friction force is zero: $\begin{aligned} & & \vec F_g + \vec F_f & \stackrel{!}{=} 0 \\ \Rightarrow & & - m g + c \rho A v_\infty^2 &= 0 \\ \Rightarrow & & v_\infty &= \sqrt{\frac{mg}{c \rho A}} \propto \frac{1}{\sqrt{A}} \end{aligned}$ When opening the parachute, only the cross-sectional area $A$ changes, so that will be able to write $\begin{aligned} & & v_1 \sqrt{A_1} &= v_2 \sqrt{A_2} \\ \Rightarrow & & v_2 &= \sqrt{\frac{A_1}{A_2}} v_1 \\ & & &\approx \sqrt{\frac{1}{100}} \cdot 50 \,\frac{\text{m}}{\text{s}} \\ & & &= 5 \,\frac{\text{m}}{\text{s}} \end{aligned}$ In fact, the frictional force depends on the shape of the object. The equation for the drag is according to the textbook: $F_f = \frac{1}{2} c_w \rho A v^2$ with the dimensionless parameter $c_w$ , which is for a parachute $c_ {w2} \approx 1.33$ and for a human $c_ {w1} \approx 0.78$ . Taking the $c_w$ -value into consideration, we therefore obtain the final velocity $v_2 = \sqrt{\frac{c_{w1} A_1}{c_{w2} A_2}} v_1 \approx 3.8 \,\frac{\text{m}}{\text{s}}$ which is a bit smaller as our previous value.

For air, where the viscosity is relatively low, the drag force varies as $F=cAv^2$ , where $c$ depends on the shape of the object, $A$ is the cross section facing the air stream, and $v$ is the velocity. Without the parachute the steady state (limiting) velocity obeys $F_1=mg=c_1A_1v_1^2$ . With the parachute we have $F_2=mg=c_2A_2v_2^2$ . For the estimation we assume $c_1=c_2$ . Dividing the two equations yields

$1=\frac{A_1v_1^2}{A_2v_2^2}$

or

$v_2=v_1 \sqrt{\frac{A_1}{A_2}}= 50\sqrt{\frac{1}{100}}=5m/s$

The equation for the air resistance acting on an object is given by $\frac{pCAv^{2}}{2}$ , where p is the air density, C is the drag coefficient and v is the velocity. It says in the assumptions to ignore the effect shape has on air resistantance, and since this is basically what the drag coefficient is, we leave it out of the formula. At first we substitute our given values for the skydiver with out a parachute:

$F_{air}$ = $\frac{p \times 1 \times 50^{2}}{2}$ = $1250p$

We are not given the air density, but this is ok. We then substitute the values for the skydiver with his parachute in use:

$F_{air}$ = $\frac{p \times 100 \times v^{2}}{2}$ = $50pv^{2}$

Now because the total air resistance, in a way, does not change (instead others parameters do, like surface area), we can set these two answers equal to each other:

1250p = $50pv^{2}$ , which yields v = 5.

I just multiplied velocity proposed by 3.6 tout obtain the corresponding velocity in km/h. The results were aller strange: 5km/h is too slow and 57 is too fast so 18 km/h seemed the only remaining possibility (it corresponds to 5 m/s)