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Chapter 2: Problem 10

A woman bails out of an airplane at an altitude of 10,000 \(\mathrm{ft}\), falls freely for \(20 \mathrm{~s}\), then opens her parachute. How long will it take her to reach the ground? Assume linear air resistance \(\rho v \mathrm{ft} / \mathrm{s}^{2}\), taking \(\rho=0.15\) without the parachute and \(\rho=1.5\) with the parachute. (Suggestion: First determine her height above the ground and velocity when the parachute opens.)

Short Answer

Expert verified
The woman reaches the ground in approximately 153 seconds.

Step by step solution

01

Understand the Problem

The problem describes a skydiver jumping from an altitude of 10,000 ft, falling freely for 20 seconds before deploying a parachute. We need to calculate the total time taken to reach the ground considering the changes in air resistance.
02

Calculate Free Fall Descent

During free fall, the only forces acting on the woman are gravity and air resistance. Gravity provides a constant acceleration of 32 ft/s虏 downward. Air resistance is given by the formula \(-\rho v)\, ft/s^2\), where \(\rho = 0.15\). The velocity at any time \(t\) is given by \[ v(t) = \frac{g}{\rho}(1 - e^{-\rho t}) \]where \(g=32\, \text{ft/s}^2.\)After 20 seconds, the terminal velocity is given by substituting \(t=20\) into the velocity equation.
03

Calculate Height after Free Fall

Using the velocity function, calculate the height reached after 20 seconds. The position as a function of time is given by: \[ s(t) = \frac{g}{\rho^2}(\rho t - 1 + e^{-\rho t}) \]Substitute \(t=20\) to find the height.
04

Conditions with Parachute Open

Once the parachute is deployed, the air resistance increases to \(\rho = 1.5\). The new velocity function becomes: \[ v(t) = v_{20} e^{-\rho t'} + \frac{g}{\rho}(1-e^{-\rho t'}) \]where \(v_{20}\) is the velocity at \(t=20\), and \(t'\) is the time after the parachute is opened.
05

Calculate Additional Time with Parachute

With the velocity function and the remaining height (10,000 ft minus the fall distance), integrate using: \[ s(t') = s_{20} - \int_0^{T} v(t') dt' \]Solve for \(T\), the time to reach the ground after deploying the parachute.
06

Total Time to Reach Ground

The total time is the sum of time spent in free fall (20 seconds) and time spent with the parachute open \(T\). Calculate these and sum for the complete descent duration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Air Resistance
Linear air resistance is a force that opposes the motion of an object moving through air. This resistance is directly proportional to the velocity of the object. In this scenario, the formula for air resistance is given by \(-\rho v\ \, ext{ft}/ ext{s}^2\), where \(\rho\) is a constant that varies based on whether the parachute is deployed or not.
Without the parachute, the constant \(\rho = 0.15\), reflecting a lower air resistance, as the woman falls faster.
With the parachute, \(\rho = 1.5\), indicating a significant increase in resistance, slowing her descent dramatically.
  • Air resistance acts opposite to gravity.
  • Forces at play affect the speed and eventual terminal velocity.
  • The change in resistance after the parachute opens is significant and deliberate for safe landing.
Understanding air resistance helps in predicting how quickly an object will fall and how it can be slowed down effectively using devices like parachutes.
Terminal Velocity
Terminal velocity is the steady speed an object reaches when the force of air resistance equals the force of gravity acting on it. At this point, the net acceleration of the object is zero, and it falls at a constant speed.
For the first 20 seconds of the skydiver's fall, gravity and the initial level of air resistance determine her terminal velocity. The velocity equation \[ v(t) = \frac{g}{\rho}(1 - e^{-\rho t}) \] describes how her velocity approaches terminal velocity over time.
Key points about terminal velocity include:
  • It depends on the air resistance constant \(\rho\).
  • The higher the air resistance, the lower the terminal velocity.
  • Initial terminal velocity is reached in the absence of a parachute due to lower air resistance.
As her parachute opens, the terminal velocity rapidly decreases due to increased air resistance, ensuring a safer and more controlled descent.
Free Fall Dynamics
Free fall dynamics are essential in understanding the motion of falling objects under gravity's influence. During free fall, the dominant force is gravity, which accelerates the object downwards at about \(32 \, \text{ft/s}^2\).
In this problem, free fall describes the first 20 seconds. The formula to calculate the velocity, accounting for air resistance, is important in determining her velocity and increasing descent time.
  • Gravity acts constantly, pulling the skydiver down.
  • Air resistance modifies this motion, reducing acceleration.
  • After 20 seconds, her velocity helps determine the next phase of descent with a parachute.
Understanding how these forces interact is critical in analyzing skydiving dynamics, especially in calculating how long it might take before reaching the ground safely.
Parachute Descent
Parachute descent drastically alters the descent dynamics, managed by the equation once the parachute has opened: \[ v(t) = v_{20} e^{-\rho t'} + \frac{g}{\rho}(1-e^{-\rho t'}) \]
With the parachute open, the air resistance is very high \(\rho = 1.5\), which significantly reduces her fall speed.
This stage begins 20 seconds into her jump, changing from a rapid descent to a slowed and controlled one.
Key aspects include:
  • Parachute increases \(\rho\) for more air resistance.
  • The velocity equation revisions reflect increased air resistance.
  • Calculating time \(T\) to ground involves integrating the altered velocity equation, accounting for height remaining.
Parachute descent ensures safety by reducing speed significantly, allowing precise calculations of descent time remaining and ensuring the skydiver lands safely.

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Most popular questions from this chapter

A hand-held calculator will suffice for Problems 1 through 10, where an initial value problem and its exact solution are given. Apply the improved Euler method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.1 .\) Construct a table showing four-decimal-place values of the approximate solution and actual solution at the points \(x=0.1,0.2,0.3\), \(0.4,0.5\) $$ y^{\prime}=-2 x y, y(0)=2 ; y(x)=2 e^{-x^{2}} $$

As in Problem 25 of Section \(2.4\), you bail out of a helicopter and immediately open your parachute, so your downward velocity satisfies the initial value problem $$ \frac{d v}{d t}=32-1.6 v, \quad v(0)=0 $$ (with \(t\) in seconds and \(v\) in \(\mathrm{ft} / \mathrm{s}\) ). Use the improved Euler method with a programmable calculator or computer to approximate the solution for \(0 \leqq t \leqq 2\), first with step size \(h=0.01\) and then with \(h=0.005\), rounding off approximate \(v\) -values to three decimal places. What percentage of the limiting velocity \(20 \mathrm{ft} / \mathrm{s}\) has been attained after 1 second? After 2 seconds?

(a) Show that if a projectile is launched straight upward from the surface of the earth with initial velocity \(v_{0}\) less than escape velocity \(\sqrt{2 G M / R}\), then the maximum distance from the center of the earth attained by the projectile is $$ r_{\max }=\frac{2 G M R}{2 G M-R v_{0}^{2}} $$ where \(M\) and \(R\) are the mass and radius of the earth, respectively. (b) With what initial velocity \(v_{0}\) must such a projectile be launched to yield a maximum altitude of 100 kilometers above the surface of the earth? (c) Find the maximum distance from the center of the earth, expressed in terms of earth radii, attained by a projectile launched from the surface of the earth with \(90 \%\) of escape velocity.

Example 4 dealt with the case \(4 h>k M^{2}\) in the equation \(d x / d t=k x(M-x)-h\) that describes constant-rate harvesting of a logistic population. Problems deal with the other cases. If \(4 h>k M^{2}\), show that \(x(t)=0\) after a finite period of time, so the lake is fished out (whatever the initial population). [Suggestion: Complete the square to rewrite the differential equation in the form \(d x / d t=-k\left[(x-a)^{2}+b^{2}\right]\). Then solve explicitly by separation of variables.] The results of this and the previous problem (together with Example 4) show that \(h=\frac{1}{4} k M^{2}\) is a critical harvesting rate for a logistic population. At any lesser harvesting rate the population approaches a limiting population \(N\) that is less than \(M\) (why?), whereas at any greater harvesting rate the population reaches extinction.

Apply Euler's method with successively smaller step sizes on the interval \([0,2]\) to verify empirically that the solution of the initial value problem $$ \frac{d y}{d x}=x^{2}+y^{2}, \quad y(0)=0 $$ has a vertical asymptote near \(x=2.003147 .\) (Contrast this with Example 2 , in which \(y(0)=1\).)
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