Question

For the free-falling parachutist with linear drag, assume a first jumper is 80 kg and has a drag
coefficient of 12 kg/s. If a second jumper has a drag coefficient of 15 kg/s and a mass of 93 kg,
how long will it take the second jumper to reach the same velocity as the first jumper reached in 10 s?

Answer 1

Answer on Question#38423 – Engineering - Other

For the free-falling parachutist with linear drag, assume a first jumper is $80\,\mathrm{kg}$ and has a drag coefficient of $12\,\mathrm{kg/s}$. If a second jumper has a drag coefficient of $15\,\mathrm{kg/s}$ and a mass of $93\,\mathrm{kg}$, how long will it take the second jumper to reach the same velocity as the first jumper reached in 10 s?

Solution:

Taking downward to be positive, we have:

$$\frac{\mathrm{d}v}{\mathrm{d}t} = \frac{F}{m} = \frac{\mathrm{mg} - kv}{m} = g - \left(\frac{k}{m}\right) v$$

Solving the separable equation, we have

$$t + C = -m \frac{\ln \left(g - \frac{kv}{m}\right)}{k}$$

So that

$$g - \frac{kv}{m} = A e^{-\frac{kt}{m}}, \text{ where } v = 0 \text{ when } t = 0, \text{ so } A = g.$$ $$v = \frac{\mathrm{mg} \left(1 - e^{-\frac{kt}{m}}\right)}{k}$$

Therefore, after 10 seconds, the first jumper attains a velocity of

$$v = \frac{80\,\mathrm{kg} \cdot 9.8\,\frac{\mathrm{N}}{\mathrm{kg}} \cdot \left(1 - e^{-\frac{12\,\mathrm{kg} \cdot 10\,\mathrm{s}}{80\,\mathrm{kg}}}\right)}{12\,\frac{\mathrm{kg}}{\mathrm{s}}} = 50.8\,\frac{\mathrm{m}}{\mathrm{s}}$$

The amount of time it takes the second jumper is

$$t = -m \cdot \frac{\ln \left(1 - \frac{kv}{\mathrm{mg}}\right)}{k} = -93\,\mathrm{kg} \cdot \frac{\ln \left(1 - \frac{15\,\frac{\mathrm{kg}}{\mathrm{s}} \cdot 50.8\,\frac{\mathrm{m}}{\mathrm{s}}}{93\,\mathrm{kg} \cdot 9.8\,\frac{\mathrm{N}}{\mathrm{kg}}}\right)}{15\,\frac{\mathrm{kg}}{\mathrm{s}}} = 11.2\,\mathrm{s},$$

Answer: the amount of time it takes the second jumper is $11.2\,\mathrm{s}$