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Adam's calculation was apparently based on v = 32 t, as 120 mph = 176 ft/sec, and 176/32 = 5.5.
The height was then calculated from s = 16t^2 = 16 (5.5)^2 = 484 feet, so that 500 feet would be sufficient.
However, that assumes that air resistance begins only when terminal velocity is reached. Actually it begins as soon as the body starts to fall and increases as the body approaches terminal velocity asymptotically.
As the acceleration due to air resistance varies with the square of the velocity, we can find the constant of proportionality, c, from the knowledge of what the terminal velocity is:
32.174 = c (176)^2
leading to c = 1/962.76, so that the total downward acceleration acting on the body at a given moment is 32.174 - v^2/962.76 ft/sec^2.
The integration of the motion can be accomplished analytically, but it would be easier to do numerical integration. Analytically, at any given time, from being dropped, assuming g = 32.174 ft/sec^2:
velocity: v = sqrt(32.174/c) tanh (t * sqrt(32.174 * c))
distance: s = ln(cosh(t * sqrt(32.174 * c))) / c
Using a computer program to integrate this numerically, we get a table, with tabulated entries every second, but with the integration interval being 0.0001 seconds for accuracy. In addition to the 1-second tabulations, one entry is provided as soon as the drop has exceeded 500 feet--the proposed distance at which terminal velocity was presumed to have been "reached":
-------Numerical------- ----Analytic---
sec ft/s mph dist.dropped ft/s distance
1.00 31.82 21.70 16.00 31.82 16.00
2.00 61.63 42.02 62.96 61.63 62.96
3.00 87.88 59.92 138.06 87.88 138.06
4.00 109.79 74.86 237.27 109.79 237.27
5.00 127.26 86.77 356.16 127.26 356.16
6.00 140.69 95.92 490.44 140.69 490.44
6.07 141.47 96.46 500.01 141.47 500.01
7.00 150.73 102.77 636.40 150.72 636.40
8.00 158.07 107.78 791.00 158.07 790.99
9.00 163.36 111.38 951.86 163.36 951.86
10.00 167.14 113.96 1117.22 167.14 1117.22
11.00 169.80 115.77 1285.77 169.80 1285.76
12.00 171.68 117.05 1456.56 171.68 1456.56
13.00 172.99 117.95 1628.94 172.99 1628.93
14.00 173.91 118.57 1802.41 173.91 1802.41
15.00 174.54 119.01 1976.66 174.54 1976.65
16.00 174.99 119.31 2151.44 174.99 2151.43
17.00 175.30 119.52 2326.59 175.30 2326.58
18.00 175.51 119.67 2502.00 175.51 2502.00
19.00 175.66 119.77 2677.59 175.66 2677.59
20.00 175.77 119.84 2853.31 175.77 2853.30
21.00 175.84 119.89 3029.11 175.84 3029.11
22.00 175.89 119.92 3204.98 175.89 3204.97
23.00 175.92 119.95 3380.88 175.92 3380.88
24.00 175.95 119.96 3556.82 175.95 3556.81
25.00 175.96 119.97 3732.77 175.96 3732.77
26.00 175.97 119.98 3908.74 175.97 3908.73
27.00 175.98 119.99 4084.72 175.98 4084.71
28.00 175.99 119.99 4260.70 175.99 4260.70
29.00 175.99 119.99 4436.69 175.99 4436.69
30.00 175.99 120.00 4612.68 175.99 4612.68
31.00 176.00 120.00 4788.68 176.00 4788.67
32.00 176.00 120.00 4964.68 176.00 4964.67
This indicates that after dropping 500 feet (occurring just over 6 seconds, rather than 5.5), the body is going only 96 mph, rather than the 120 mph terminal velocity. Most would say that with a fall from 1973 feet, at 119 mph, that would be "close enough" to terminal velocity, and that is the answer to the second part.
DEFDBL A-Z
incr = .0001
v = 0
c = 32.174# / (176 * 176): PRINT 1 / c
DO
acc = 32.174 - c * v * v
vNew = acc * incr + v
s = s + incr * (vNew + v) / 2
v = vNew
t = t + incr
ct = ct + 1
IF ct MOD 10000 = 0 OR s > 500 AND fsw = 0 THEN
argH = t * SQR(32.174 * c)
vAn = SQR(32.174 / c) * ((EXP(argH) - EXP(-argH)) / (EXP(argH) + EXP(-argH)))
sAn = LOG((EXP(argH) + EXP(-argH)) / 2) / c
PRINT USING "###.## ####.## ####.## #####.## ####.## #####.##"; t; v; v * 3600 / 5280; s; vAn; sAn
IF s > 500 THEN fsw = 1
END IF
LOOP UNTIL s > 5000 AND ct MOD 10 = 0
The analytical solution has 500 feet reached at t=6.0678 seconds, traveling 141.6669 ft/sec or 96.4547 mph, found by iteration in this extension to the above program:
tb = 5: te = 8
DO
t = (tb + te) / 2
argH = t * SQR(32.174 * c)
vAn = SQR(32.174 / c) * ((EXP(argH) - EXP(-argH)) / (EXP(argH) + EXP(-argH)))
sAn = LOG((EXP(argH) + EXP(-argH)) / 2) / c
IF sAn < 500 THEN tb = t
IF sAn > 500 THEN te = t
LOOP UNTIL ABS(sAn - 500) < .000001
PRINT t, vAn
A similar iteration for the analytic solution finds 119 mph (174.533... ft/sec) reached at 14.98 sec., at 1972.96 feet.
tb = 5: te = 20
DO
t = (tb + te) / 2
argH = t * SQR(32.174 * c)
vAn = SQR(32.174 / c) * ((EXP(argH) - EXP(-argH)) / (EXP(argH) + EXP(-argH)))
sAn = LOG((EXP(argH) + EXP(-argH)) / 2) / c
IF vAn < 119# * 5280 / 3600 THEN tb = t
IF vAn > 119# * 5280 / 3600 THEN te = t
LOOP UNTIL ABS(vAn - 119# * 5280 / 3600) < .000001
PRINT t, vAn, sAn
Alternatively to the binary search method of solving for t, formulas exist for inverse hyperbolic functions:
invSinh(x) = ln(x + sqrt(x^2 + 1))
invCosh(x) = ln(x + sqrt(x-1) * sqrt(x+1))
invTanh(x) = (ln(1 + x) - ln(1 - x)) / 2
so to find how fast Buster is going at 500 feet, first find the time taken, via
500 = ln(cosh(t * sqrt(32.174 * c))) / c
.51934 = ln(cosh(t * sqrt(32.174 * c)))
1.680918 = cosh(t * sqrt(32.174 * c))
1.10923 = t * sqrt(32.174 / 962.76)
t = 6.0677576
(intermediate results carried to more places than shown)
then velocity calculated from v = sqrt(32.174/c) tanh (t * sqrt(32.174 * c))
Similar calculation can find the time from v = 119 * 5280 / 3600, and then the distance from s = ln(cosh(t * sqrt(32.174 * c))) / c.
Now that we have established that .0001 is an adequately small interval of integration, it may be desired to take into consideration the changed atmospheric pressure and temperature at increased altitude. A rule of thumb is that barometric pressure is 1 inch lower per 1000 feet of altitude near the earth's surface. Temperature varies at about 3 degrees Celsius per thousand feet.
By the perfect gas law, the density of the air would be proportional to the pressure divided by the temperature.
If we assume the terminal velocity was measured at sea level at 30 inches of mercury and at 288 degrees Kelvin, the density would be (1-.023*a/1000), where a is the altitude, and air resistance would be that much smaller in proportion. The changed lines in the program:
ht0 = 500
acc = 32.174 - c * (1 - .023 * (ht0 - s) / 1000) * v * v
results in very little change in the velocity at 500 feet:
6.06 141.57 96.53 500.00
It's still between 96 and 97 mph.
Trial and error with ht0 indicates that when falling from 1785 feet, 119 mph will be achieved by groundfall at 13.82 seconds:
13.82 174.54 119.01 1785.02
Interestingly, if we set the initial altitude to 22,000 feet, the body's velocity gets up to a maximum of about 160 mph after falling somewhat over 3000 feet in the thinner air, but reduces to just over 120 mph at the ground:
(The displayed inverval in the below table is shown as 3 seconds so as not to be too long)
3.00 91.99 62.72 141.33
6.00 161.74 110.28 529.05
9.00 203.49 138.74 1083.35
12.00 224.03 152.75 1728.74
15.00 232.16 158.29 2415.22
18.00 234.00 159.55 3115.51
21.00 232.95 158.83 3816.40
24.00 230.67 157.27 4512.00
27.00 227.94 155.41 5199.97
30.00 225.10 153.48 5879.54
33.00 222.31 151.57 6550.63
36.00 219.61 149.73 7213.48
39.00 217.02 147.97 7868.38
42.00 214.54 146.28 8515.69
45.00 212.17 144.66 9155.74
48.00 209.91 143.12 9788.84
51.00 207.74 141.64 10415.29
54.00 205.66 140.22 11035.37
57.00 203.66 138.86 11649.34
60.00 201.74 137.55 12257.43
63.00 199.90 136.29 12859.88
66.00 198.11 135.08 13456.87
69.00 196.40 133.91 14048.63
72.00 194.74 132.78 14635.31
75.00 193.13 131.68 15217.11
78.00 191.58 130.63 15794.18
81.00 190.08 129.60 16366.66
84.00 188.63 128.61 16934.72
87.00 187.22 127.65 17498.48
90.00 185.85 126.72 18058.07
93.00 184.52 125.81 18613.62
96.00 183.23 124.93 19165.24
99.00 181.98 124.07 19713.04
102.00 180.75 123.24 20257.12
105.00 179.57 122.43 20797.60
108.00 178.41 121.64 21334.55
111.00 177.28 120.87 21868.08
111.74 177.01 120.69 22000.02
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