Question

A skydiver of mass \(83.7 \mathrm{~kg}\) (including outfit and equipment) falls in the spread-eagle position, having reached terminal speed. Her drag coefficient is \(0.587,\) and her surface area that is exposed to the air stream is \(1.035 \mathrm{~m}^{2} .\) How long does it take her to fall a vertical distance of \(296.7 \mathrm{~m} ?\) (The density of air is \(\left.1.14 \mathrm{~kg} / \mathrm{m}^{3} .\right)\)

Short answer

Answer: Approximately 4.95 seconds.

Step-by-step solution

Calculate Drag Force

First, we need to calculate the force of drag experienced by the skydiver. The drag force (FD) can be found using the equation: $$ FD = \frac{1}{2} \times \rho \times v^{2} \times C_{D} \times A $$ where ρ = density of air (1.14 kg/m³), v = terminal speed of the skydiver, CD = drag coefficient (0.587), A = exposed surface area (1.035 m²). Since the skydiver has reached terminal speed, the force of gravity (FG) is equal to the drag force (FD), so: $$ FG = FD $$ The force of gravity can be calculated using the equation: $$ FG = m \times g $$ where m = mass of the skydiver (83.7 kg), g = acceleration due to gravity (approximately 9.81 m/s²).

Calculate Terminal Speed

Now we can set the drag force equal to the force of gravity and solve for the terminal speed (v): $$ m \times g = \frac{1}{2} \times \rho \times v^{2} \times C_{D} \times A $$ Plug in the given values: $$ 83.7 \times 9.81 = \frac{1}{2} \times 1.14 \times v^{2} \times 0.587 \times 1.035 $$ Solve for v: $$ v = \sqrt{\frac{2 \times 83.7 \times 9.81}{1.14 \times 0.587 \times 1.035}} \approx 59.89 \, \text{m/s} $$

Calculate Time to Fall Given Distance

Now that we have the terminal speed of the skydiver, we can use it to find the time it takes to fall a vertical distance of 296.7 m. Using the equation for distance as a function of time and constant speed: $$ d = vt $$ where d = vertical distance (296.7 m), v = terminal speed (59.89 m/s), t = time to fall. Solve for t: $$ t = \frac{d}{v} = \frac{296.7}{59.89} \approx 4.95 \, \text{s} $$ So, it takes the skydiver approximately 4.95 seconds to fall a vertical distance of 296.7 meters.