please help me out. speed = 145.295 is correct. download image if you can\'t see

Question

please help me out. speed = 145.295 is correct.

download image if you can't see

Thanks!

A person leaps from an airplane 9500 feet above the ground and deploys her/his parachute after 8 seconds. Assume that the air resistance both before and after deployment of the parachute results in a deceleration proportional to the person's velocity, but with a different constant of proportionality. Before deployment, take the constant to be 0.16, and after deployment use the value 1.4. (The acceleration due to gravity is 32.2ft/sec2. Note also that the constants of proportionality given are rho = k/m, not k.) How fast is the person going when the parachute is deployed? How high off the ground is the person when the parachute is deployed? Approximately (to one decimal place) how much longer does it take the person to reach the ground (give your answer in seconds)?

Explanation / Answer

v(t) = vo e-bt/m - (mg/b)(1- e-bt/m)

where,b/m = constant of proportionality=0.16,v0=0

v=0-32.2/0.16[1-e^(-8*0.16)]=145.295 ft/sec [- implies v is in negative y direction]

............................................................................................................................................................................

v(t) = vo e-bt/m - (mg/b)(1- e-bt/m)

so,dx/dt= vo e-bt/m - (mg/b)(1- e-bt/m)

dx=vo e-bt/m - (mg/b)(1- e-bt/m) dt

Integrating we have,

x(t)=-mvo /b e-bt/m - (mg/b)(t+m/b e-bt/m)

=0-32.2/0.16[8+1/0.16 e^(-8*0.16)]=1959.71 feet....[v0=0,- implies displacement in negative y]

.................................................................................................................................................................................

after opening parashute

use x(t)=-mvo /b e-bt/m - (mg/b)(t+m/b e-bt/m) again

this time b/m=1.4

v0=145.295

x(t)=9500-1959.71 =7540.29 feet

-7540.29= -1/1.4 * 145.295 * e^(-1.4t) -1/1.4 *32.2[t+1/1.4 e^(-1.4t)]

=-103.78e^(-1.4t) -23t -16.42 e^(-1.4t)

=-120.2 e^(-1.4t) -23 t

so, 120.2 e^(-1.4t) +23 t=7540.29

hit and trial with values of t

120.2 e^(-1.4t) is negligible

we can just take,23t=7540.29

t=327.83 sec

total time =327.83+8=335.83 sec

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