Will upvote! Problem 3Consider a sky-diving control engineer (why not?), with a parachute (of course!).Let us denote by h(t) the vertical distance (height) that the sky-driver traversed (i.e., fell by) in t,after being kicked out of the airplane. The downward velocity at that point in time t is denoted byv(t). The accelerations acting on the falling control engineer at time t are the downward gravitationalacceleration g, and the upward air-drag av(t) offered by the parachute which is a function of the size ofthe parachute (a) and the downward velocity v(t). The dynamics governing the motion of the sky-diveris given by:av(t)h(t)v(t)(height)h(t) = v(t)v(t) = g - av(t)(velocity)Assume zero initial conditions, i.e., h(0) = 0, v(0) = 0, and g = 10, a = 2.a) Use the velocity equation to find out the Laplace transform of the speed of the falling sky-diver.b) Use the Laplace transform above to calculate the time-domain trajectory of the velocity.c) Apply Final Value Theorem to calculate the steady-state velocity that the sky-diver will reach(assuming there is sufficient time to reach the steady-state speed - we do not want any brokenbones, do we?)d) Use the height equation to calculate the time-domain trajectory of the fallen height.

Given Information:Velocity: Height: .Initial conditions: .To find:a) Laplace Transform of the…

Answered: (height) h(t) = v(t) (velocity) i(t) =…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Problem 3
Consider a sky-diving control engineer (why not?), with a parachute (of course!).
Let us denote by h(t) the vertical distance (height) that the sky-driver traversed (i.e., fell by) in t,
after being kicked out of the airplane. The downward velocity at that point in time t is denoted by
v(t). The accelerations acting on the falling control engineer at time t are the downward gravitational
acceleration g, and the upward air-drag av(t) offered by the parachute which is a function of the size of
the parachute (a) and the downward velocity v(t). The dynamics governing the motion of the sky-diver
is given by:
av(t)
h(t)
v(t)
(height)
h(t) = v(t)
v(t) = g - av(t)
(velocity)
Assume zero initial conditions, i.e., h(0) = 0, v(0) = 0, and g = 10, a = 2.
a) Use the velocity equation to find out the Laplace transform of the speed of the falling sky-diver.
b) Use the Laplace transform above to calculate the time-domain trajectory of the velocity.
c) Apply Final Value Theorem to calculate the steady-state velocity that the sky-diver will reach
(assuming there is sufficient time to reach the steady-state speed - we do not want any broken
bones, do we?)
d) Use the height equation to calculate the time-domain trajectory of the fallen height.

Expert Solution

Step 1: Analysis and Introduction

Given Information:

Velocity: $v with dot on top open parentheses t close parentheses equals g minus a v open parentheses t close parentheses$

Height: $h with dot on top open parentheses t close parentheses equals v open parentheses t close parentheses$ .

$g equals 10 semicolon space a equals 2$

Initial conditions: $h open parentheses 0 close parentheses equals 0 semicolon space v open parentheses 0 close parentheses equals 0$ .

To find:

a) Laplace Transform of the velocity.

b) Velocity.

c) Velocity using final value theorem.

d) Height.

Concept used:

Final Value Theorem:

The final value theorem of Laplace transform states that if $x open parentheses t close parentheses$ has the Laplace Transform as $X open parentheses s close parentheses$ ,

then $limit as x rightwards arrow infinity of x open parentheses t close parentheses equals limit as s rightwards arrow 0 of s X open parentheses s close parentheses$ .

Laplace & Inverse Laplace formula:

1) $table row cell calligraphic L to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator s plus a end fraction close parentheses end cell equals cell e to the power of negative a t end exponent end cell end table$

2) $table row cell calligraphic L to the power of negative 1 end exponent open parentheses 1 over s close parentheses end cell equals 1 end table$

3) $calligraphic L open parentheses y apostrophe open parentheses t close parentheses close parentheses equals s calligraphic L open parentheses y open parentheses t close parentheses close parentheses minus y open parentheses 0 close parentheses$

Integration formula:

$table row cell integral a space d t end cell equals cell a t plus c end cell row cell integral e to the power of negative a t end exponent d t end cell equals cell fraction numerator e to the power of negative a t end exponent over denominator negative a end fraction plus c end cell end table$

Here, $a$ is any constant and $c$ is the integrating constant.