The velocity v of a freefalling skydiver is well modeled by the differential equ
ID: 2944269 • Letter: T
Question
The velocity v of a freefalling skydiver is well modeled by the differential equationm dv/dt =mg- kv^2,
where m is the mass of the skydiver, g =9.8m/s^2 is the gravitational constant, and k is the drag coefficient determined by the position of the diver during the dive. Consider a diver of mass m =54kg (120lb) with a drag coefficient of 0.18 kg/m. Use Euler's method to determine how long it will take the diver to reach 95% of her terminal velocity after she jumps from the plane. [Hint use the formula for terminal velocity].
Explanation / Answer
m dv/dt = mg - kv2,
dv/dt = g - kv2/m
when v = terminal velocity u, dv/dt = 0, so g = ku2/m, u = (mg/k)
dv/dt = ku2/m - kv2/m
dv/(u2 - v2) = (k/m) dt
dv/(u + v) + dv/(u - v) = (2ku/m) dt
integrate,
ln(u + v) - ln(u - v) = 2kut/m + C
ln[(u + v)/(u - v)] = 2kut/m + C
when t = 0, v = 0, so C = 0
when v = 0.95 u
ln(1.95/0.05) = 2kut/m
ln(1.95/0.05) = 2kt/m * (mg/k)
t = (kg/m) * 1/2 * ln(1.95/0.05) = 0.302 s
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