To find the distance fallen by the skydiver before the parachute opens, we need to calculate the distance traveled during the first 10 seconds of free fall, where the parachute is closed.

The motion of the skydiver involves both gravitational force and air resistance. Let’s denote the relevant quantities:

• m = 82 kg (mass of the skydiver with equipment),
• g = 9.8 m/s² (acceleration due to gravity),
• k₁ = 1.65 kg/s (resistance coefficient before the parachute opens),
• t₁ = 10 s (time of free fall before the parachute opens).

The equation of motion is given by Newton's second law:

$m \frac{dv}{dt} = mg - k_1 v$

Simplifying:

$\frac{dv}{dt} = g - \frac{k_1}{m}v$

Let $a = g$ and $b = \frac{k_1}{m}$, so the equation becomes:

$\frac{dv}{dt} = a - b v$

This is a first-order linear differential equation with solution:

$v(t) = \frac{a}{b} \left(1 - e^{-b t}\right)$

Step 1: Plug in values to find $v(t)$

Here:

$a = g = 9.8 \, \text{m/s}^2, \quad b = \frac{k_1}{m} = \frac{1.65}{82} \approx 0.0201 \, \text{s}^{-1}.$

So the velocity as a function of time is:

$v(t) = \frac{9.8}{0.0201} \left(1 - e^{-0.0201 t}\right) \approx 487.56 \left(1 - e^{-0.0201 t}\right) \, \text{m/s}.$

Step 2: Find the distance fallen, $s(t)$

The distance fallen is obtained by integrating $v(t)$ with respect to $t$:

$s(t) = \int_0^t v(t) \, dt = \int_0^t 487.56 \left(1 - e^{-0.0201 t}\right) \, dt.$

Splitting the integral:

$s(t) = \int_0^t 487.56 \, dt - \int_0^t 487.56 e^{-0.0201 t} \, dt.$

1. The first term is straightforward:

$\int_0^t 487.56 \, dt = 487.56 t.$

2. For the second term:

$\int_0^t 487.56 e^{-0.0201 t} \, dt = \left[ -\frac{487.56}{0.0201} e^{-0.0201 t} \right]_0^t = -24250 (e^{-0.0201 t} - 1).$

Combining:

$s(t) = 487.56 t + 24250 (1 - e^{-0.0201 t}).$

Step 3: Calculate $s(10)$

Now, substitute $t = 10$ seconds:

$s(10) = 487.56(10) + 24250 \left(1 - e^{-0.0201(10)}\right).$

First, calculate $e^{-0.0201(10)}$:

$e^{-0.0201(10)} \approx e^{-0.201} \approx 0.8187.$

So:

$s(10) = 4875.6 + 24250 (1 - 0.8187).$

Simplify:

$s(10) = 4875.6 + 24250 (0.1813) \approx 4875.6 + 4394.625.$

$s(10) \approx 9269.8 \, \text{m}.$

Thus, the distance fallen before the parachute opens is approximately 926.98 meters.

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Questions:

1. What would the effect of a higher air resistance coefficient $k_1$ be on the distance fallen?
2. How does the time of free fall (10 seconds) influence the distance calculation?
3. Can you derive the terminal velocity from this problem?
4. What happens to the distance calculation if the mass of the skydiver changes?
5. How does varying $g$ (e.g., on another planet) alter the distance fallen?

Tip:

Always check if the air resistance term significantly affects velocity, especially for long durations of free fall.