Essay Available:
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13 pages/β3575 words
Sources:
Check Instructions
Style:
Harvard
Subject:
Mathematics & Economics
Type:
Math Problem
Language:
English (U.K.)
Document:
MS Word
Date:
Total cost:
$ 56.16
Topic:
Finding Terminal Velocity and Using Logarithms and Exponential Calculations
Math Problem Instructions:
please write in the word document, the excel work must include the function formula not the copy and paste of the numbers, and send separately. I want the distinction level and the explanation is written in the assignment brief. I also want in-text citation as well as Harvard referencing. I live in the European region so the references most work in Europe. thank you very much
Math Problem Sample Content Preview:
Q1.
*
At any given time, the velocity is given by the formula:
The given in the case are;
M = 68.1 kg
G = 9.81
C = 12.51
By inserting these into the equation, we can get the terminal velocity which is;
V(t) = [9.81(68.1)/ 12.5](1-e-(12.5/68.1)t)
= 53.39(1-e-0.18355t)
Therefore, the formula for the terminal velocity prior to opening the chute could be seen in this formula 53.39(1-e-0.18355t)
*
Time in Sec.
Velocity in m/s
0
0.00
4
27.77
8
41.09
12
47.49
16
50.56
20
52.03
24
52.74
28
53.08
32
53.24
36
53.32
40
53.36
50
53.38
54
53.39
* The terminal velocity for this is 53.39 m/s
Q2.
The root of the function f(x) = x10 – 1 by using the bisection method. In order to do this, we will start with the initial guesses a = 0 and b = 1.3.
a
b
f(a)
f(b)
c
f(c)
b-a
0
1.3
-1
12.8
0.65
...
*
At any given time, the velocity is given by the formula:
The given in the case are;
M = 68.1 kg
G = 9.81
C = 12.51
By inserting these into the equation, we can get the terminal velocity which is;
V(t) = [9.81(68.1)/ 12.5](1-e-(12.5/68.1)t)
= 53.39(1-e-0.18355t)
Therefore, the formula for the terminal velocity prior to opening the chute could be seen in this formula 53.39(1-e-0.18355t)
*
Time in Sec.
Velocity in m/s
0
0.00
4
27.77
8
41.09
12
47.49
16
50.56
20
52.03
24
52.74
28
53.08
32
53.24
36
53.32
40
53.36
50
53.38
54
53.39
* The terminal velocity for this is 53.39 m/s
Q2.
The root of the function f(x) = x10 – 1 by using the bisection method. In order to do this, we will start with the initial guesses a = 0 and b = 1.3.
a
b
f(a)
f(b)
c
f(c)
b-a
0
1.3
-1
12.8
0.65
...
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