   Chapter 3.11, Problem 54E

Chapter
Section
Textbook Problem

A model for the velocity of a falling object after time t is v ( t ) = m g k tanh ( t g k m ) where m is the mass of the object, g = 9.8 m/s2 is the acceleration due to gravity, k: is a constant, t is measured in seconds, and v in m/s.(a) Calculate the terminal velocity of the object, that is, limt→∞ v(t).(b) If a person falls from a building, the value of the constant k depends on his or her position. For a “belly-to-earth” position, k = 0.515 kg/s, but for a “feet-fist” position, k = 0.067 kg/s. If a 60-kg person falls in belly-to-earth position, what is the terminal velocity? What about feet-first?Source: L. Long et at, “How Terminals Terminal Velocity?” American Mathematical Monthly 113 (2006): 752–55.

(a)

To determine

To find: The terminal velocity of the objects. That is, limtv(t).

Explanation

Given:

The velocity of a falling object after time t is, v(t)=mgktanh(tgkm).

where, m is mass of the object, g is acceleration due to gravity and k is constant.

Calculation:

Consider the velocity of the objects is v(t)=mgktanh(tgkm).

Here the time t is an independent variable but k, g and m is fixed value of the objects and it is independents on the time variable.

Thus, label p=mgk and q=gkm.

v(t)=mgktanh(tgkm)

Substitute p=mgk and q=gkm,

v(t)=ptanh(tq)

It required that the terminal velocity of the objects.

That is, obtain the velocity after time t when time is larger and large.

limtv(t)=limtptanh(tq)=plimttanh(tq)

Put x=tq then x is approach to infinitely as time t approaches to infinitely.

Obtain the value limxtanhx

(b)

To determine

To find: The terminal velocity of 60 kg person falls in belly to earth and the terminal velocity of 60 kg person falls in feet first.

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