Question

A $75kg$skydiver can be modeled as a rectangular “box” with dimensions $20cm*40cm*180cm.$What is his terminal speed if he falls feet first? Use $0.8$for the drag coefficient.

Short answer

The terminal speed if he falls feet first is$136.72m/sec.$

Step-by-step solution

Given information.

$\text{Mass(m)=}75kg,\text{dimensions of the box shaped body =}20cm*40cm*180cm,\text{drag coefficient(C)=}0.8.$

Calculation

The formula for terminal velocity is given by:

$v=\sqrt{\frac{2mg}{\rho ac}}.$

here,

$\mathrm{m}=$mass of the body

$\mathrm{g}=$gravitational acceleration

$\rho =$density of air

$a$= Projected area

c = Drag Constant

The density of air islocalid="1648476929997" $1.23\mathrm{kg}/{\mathrm{m}}^3.$

Projected area :

$$\begin{gathered} \left(a\right)=20*40. \\ =800{\mathrm{cm}}^2. \\ =0.08{\mathrm{m}}^2. \end{gathered}$$

Continuation of calculation

Putting the values we get,

$$\begin{gathered} v=\sqrt{\frac{2*\left(75kg\right)*(9.81m/s^2)}{(1.23kg/m^3)*(0.08m^2)*(0.8)}}. \\ v=\sqrt{18692.8354m^2/s^2}. \\ v=136.72\mathrm{m}/\sec . \end{gathered}$$

Therefore, the terminal velocity is$136.72m/sec.$

Final answer

The terminal speed if he falls feet first is$136.72m/sec.$