Question

this project discovers the free-falling velocity of skydivers before the parachutes are opened using the laws of physics and calculus. you can ignore the wind in the horizontal direction. let m be the mass of a skydiver and the equipment, g be the acceleration due to gravity. the free-falling velocity of a skydiver, v(t), increases with time. the force due to the air resistance is correlated with the velocity, that is, Fr=kv^2, where k>0 if called the drag constant related to the amount of air resistance

MAN Differential Equations Project (optional), due on Wednesday '1,2019 This project discovers the free-falling velocity of skydivers before the parachutes are opened using the laws of physics and Calculus. You can ignore the wind in the horizontal direction. Let m be the mass of a skydiver and the equipment, g be the acceleration due to gravity. The free-falling velocity of a skydiver, v(t), increases with time. The force due to the air resistance is correlated with the velocity, that is, = kv 2, where k>0 is called the drag constant related to the amount of air resistance (a) Draw a free-body diagram showing all forces acting on the skydiver in the vertical direction. (b) Develop a differential equation to model the free-falling velocity of the skydiver. Also find the initial velocity v(0). (c) Now you will solve the differential equation in part (b). Note that your solution for v shall be in term of t. Use the initial velocity v(0) to find the constant term C. (d) Based on your solution in part (c) to find the limit of v(t) as t → oo. This velocity is called terminal velocity V (e) An experimental study confirmed that a good model gives a good quantitative fit to the data of human skydivers. Six men were dropped from altitudes varying from 10,600 to 31,400 feet to a terminal altitude of 2,100 feet, at which they opened their parachutes. The long free fall from 31,400 to 2,100 feet took 116 seconds. The average of the man and their equipment was 216.2 pounds. In these unites, g 32.2 ft/sec?. Compute the average velocity vavg Now you will use the data give in part (e) to estimate the terminal velocity V and the value of drag constant k (accurate to two decimal places) First you need to find the exact formula for s(t), the disant fallen, using the identity t- -v() and s(0)0, where () is known from your result in part (c). Then find V using t 116 sec, s (0 (31,400 -2,100) ft, and g 32.2 ft/sec2.

Answer 1

Solution:

a)

Drag Weight

$\text{b)}$

By Newton's laws of motion,

(0)0 ma- ku

du dt = mg-h?

dv = dt mg - kv

$\frac{\mathrm{d} v}{g-\frac{k}{m}v^2}=\mathrm{d} t$

dt V)

m In Vgkm

$\text{c)}$

$\text{applying initial condition we get, } C=0$

kv TL m In tVgknm vgkm

$\dfrac{m\ln\left(\frac{\left|kv-\sqrt{gkm}\right|}{\left|kv+\sqrt{gkm}\right|}\right)}{2\sqrt{gkm}}=-t$

$\ln\left(\frac{\left|kv-\sqrt{gkm}\right|}{\left|kv+\sqrt{gkm}\right|}\right)=-2\sqrt{\frac{kg}{m}}t$

$\frac{\left|kv-\sqrt{gkm}\right|}{\left|kv+\sqrt{gkm}\right|}=e^{-2\sqrt{\frac{kg}{m}}t}$

$\frac{kv-\sqrt{gkm}}{kv+\sqrt{gkm}}=e^{-2\sqrt{\frac{kg}{m}}t}$

$kv-\sqrt{gkm}=\left ( kv+\sqrt{gkm} \right )e^{-2\sqrt{\frac{kg}{m}}t}$

$kv-kve^{-2\sqrt{\frac{kg}{m}}t}=\sqrt{gkm} e^{-2\sqrt{\frac{kg}{m}}t}+\sqrt{gkm}$

$kv\left (1-e^{-2\sqrt{\frac{kg}{m}}t} \right )=\sqrt{gkm}\left (1+e^{-2\sqrt{\frac{kg}{m}}t} \right )$

$kv=\sqrt{gkm}\frac{\left (1+e^{-2\sqrt{\frac{kg}{m}}t} \right )}{\left (1-e^{-2\sqrt{\frac{kg}{m}}t} \right )}$

$v(t)=\sqrt{\frac{mg}{k}}\frac{\left (1+e^{-2\sqrt{\frac{kg}{m}}t} \right )}{\left (1-e^{-2\sqrt{\frac{kg}{m}}t} \right )}$

d)

$\text{as t tends to inifinity the velocity of the parachute will be,}$

$v(t)|_{t \to \infty}=\lim_{t \to \infty}\sqrt{\frac{mg}{k}}\frac{\left (1+e^{-2\sqrt{\frac{kg}{m}}t} \right )}{\left (1-e^{-2\sqrt{\frac{kg}{m}}t} \right )}$

$v(t)|_{t \to \infty}=\lim_{t \to \infty}\sqrt{\frac{mg}{k}}\frac{\left ( 1+0 \right )}{\left ( 1-0 \right )}$

$v(t)|_{t \to \infty}=\sqrt{\frac{mg}{k}}$