Question

A skydiver weighing 176 lbs drops from a plane that is at an altitude of 5000 f Assume that g = 32 ft/s². mg=176. (Note: the Diff. equ is mv'=-mg-cv, which can be written as v'=-g-cv/m. The general solution of this is v=-mg/c+Ae^(-ct/m), where A is an arbitrary constant and m=176/g=11/2)

a) Before the parachute opens, the forces on the skydiver are gravity and a drag force Fr= -cv. Assuming v(0) = 0, write down the IVP for v, and then find the solution. (Please show the steps using the above info.)

b) It is claimed that the terminal velocity of a person falling is -120 mph. Use this to determine c.

c) If the parachute is opened after 10s of free fall, what is the speed of the skydiver when it opens?

d) Find the distance the skydiver falls before the parachute opens

Answer 1

Step-by-step explanation:

a) The general solution is given to be:

v = -mg/c + Ae^(-ct/m)

To derive this, start with the differential equation:

m dv/dt = -mg − cv

Separate the variables:

dv / (mg + cv) = -1/m dt

c dv / (mg + cv) = -c/m dt

Integrate:

ln(mg + cv) = -c/m t + A

mg + cv = e^(-c/m t + A)

mg + cv = Ae^(-c/m t)

cv = -mg + Ae^(c/m t)

v = -mg/c + Ae^(c/m t)

*Notice that A is an arbitrary constant, so multiplying it by another constant is still an arbitrary constant.

At t=0, v=0:

0 = -mg/c + A

A = mg/c

Substitute:

v = -mg/c + mg/c e^(-ct/m)

v = -mg/c (1 − e^(-ct/m))

Given that mg = 176 and m = 11/2:

v = -176/c (1 − e^(-2ct/11))

b) The terminal velocity occurs as t approaches infinity.

-120 = -176/c (1 − 0)

c = 176/120

c = 22/15

c) Plugging in the value for c, the velocity equation is:

v = -120 (1 − e^(-4t/15))

At t = 10, the velocity is:

v = -120 (1 − e^(-40/15))

v = -111.66 ft/s

d) To find the distance, integrate from t=0 to t=10.

d = ∫₀¹⁰ v dt

d = ∫₀¹⁰ -120 (1 − e^(-4t/15)) dt

d = ∫₀¹⁰ (-120 + 120 e^(-4t/15)) dt

d = ∫₀¹⁰ (-120 − 450 × -4/15 e^(-4t/15)) dt

d = (-120t − 450 e^(-4t/15)) |₀¹⁰

d = (-1200 − 450 e^(-40/15)) − (0 − 450)

d = -750 − 450 e^(-8/3)

d = -781.27 ft