(1 point) Math 216 Homework webHW3, Problem 6 A person leaps from an airplane 55

Question

(1 point) Math 216 Homework webHW3, Problem 6 A person leaps from an airplane 5500 feet above the ground and deploys herhis parachute after 6 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 0.1, and after deployment use the value 1 6 The acceleration due to gravity is 322t sec2 Note also that the constants of proportionaity gi en are notk) How fast is the person going when the parachute is deployed? speed = 145.2826532 How high off the ground is the person when the parachute is deployed? height-5020.826532 Approximately (to one decimal place) how much longer does it take the person to reach the ground (give your answer in seconds)? time

Explanation / Answer

here force acting on the person will
due to gravity = mg and
due to resistance = kx
now before the parachute is opened k/m = 0.1 ad after it is opened k/m=1.6

net force on the person ma = mg - kv
hence a = (mg - kv)/m
= g - (k/m)v
lets consider the case of before opening
acceleration a =dv/dt
therefore
dv/dt = g - (k/m)v
this is simple integration which on integrating will give
ln(g - kv/m) = t + c (1) ... (c is integration constant)
giving condition that at the instance the man falls i.e at t=0, v=0
we get c=-ln(g)
substituting this value in equation (1) and then putting given data and solving for v we get

v = 145.28 ft/(sec^2) ... this is answer for part 1

now we know a=d^2s/dt^2. which is second order form.
solving step by step this differential equation we get
at = ds/dt + q
and again integrating we get
0.5at^2 = s +qt + p (2)... (q and p are integration constants)
putting the boundary conditions
at t=0 , s=0
and at t=0 , ds/dt = 0

we get the values of q and p,
putting these values of q and p in equation (2) and also putting data we get
s = 5020.82 ft ...this is answer for part 2

now again a=vdv/ds
a = g - (k/m)v as we did previously but here k/m = 1.4
solving again like we did in the 2nd part and finding the integration constants

vdv/ds = g - kv/m

dv/ds = g/v - k/m
integrating above we get
s = g/(g-kv/m) - m/k + r ...... r is integration constant
putting boundary condition from what we obtained from 1st part
i.e. at s=5020.82 , v=145.28
and solving we get value of r.
once we get r we can put v = ds/dt and again find s in terms of t. now integrating that equation from s = 5020.82 ft to s = 5500 ft and t from zero to t
we get an equation in terms of t.
solve that equation and we can get the time required by that person to land after teh parachute was deployed.

Hope this will help !

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