Incorrect. A skydiver weighing 270 lb (including equipment) falls vertically dow

Question

Incorrect. A skydiver weighing 270 lb (including equipment) falls vertically downward from an altitude of 6000 ft and opens the parachute after 12 s of free fall. Assume that the force of air resistance, which is directed opposite to the velocity, is 0.79Ivl when the parachute is closed and 12lvI when the parachute is open, where the velocity v is measured in ft/s. use g = 32 ft/s. Round your answers to two decimal places (a) Find the speed of the skydiver when the parachute opens 271.31 y(12) = ft/s (b) Find the distance fallen before the parachute opens x(12) (c) What is the limiting velocity Vi after the parachute opens? 103.21

Explanation / Answer

given

weight of skydiver, m = 270 lb

altitude, h = 6000ft

time t = 12 s

Fr = 0.79|v| ( where v is the velocity)

for open parachute

fr = 12|v|

a. let the speed of the diver when the parachute opens be u

then

at time t < 12 s

speed = v

from force balance

mg - Fr = mdv/dt

270*32 + 0.79v = 270dv/dt

dt/270 = dv/(8640 + 0.79v)

integrating from v = 0 to v = u

t/270 = ln(1 + 0.79u/8640)

t = 12 s

hence

u = 497 ft/s

b. distance fallen when the parachute opens = s

at distance y

(8640 + 0.79v) = 270dv*dy/dt*dy = 270*vdv/dy

dy = 270vdv/(8640 + 0.79v) = [341.772151(8640 + 0.79v)-2952911.3924]dv/(8640 + 0.79v)

dy = 341.772dv - 2952911.3924dv/(8640 + 0.79v)

integrating

y = 169860.684 - 131230.31626 = 38630.3677 ft

c. after the parachute opens limiting speed = v'

270*32 = 12*v'

v' = 720 ft/s

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