skydiving (differential equations)?

Question

A skydiver weights 125 pounds, and parachute another 35 pounds. After exiting from plane at an altitude of 15000 feet she waits 15 seconds and opens parachute.The positive direction down. value of k=0.5 for free fall and k=10 after the parachute is opened. Assume his initial velocity on leaving the plane is 0. what his velocity and how far has she traveled 20 seconds after leaving the plane? how does his velocity at 20 seconds compare with his terminal velocity? how long does it take to reach the ground.[hint think in terms of two distinct IVPs]

please somebody
I have lost myself in there
can someone get solution in steps
thank you very much

Answer 1

m dv/dt = mg - kv ..................... mass * acceleration = net force

dv/dt + k/m v = g

d/dt [ e^(kt/m) v ] = g e^(kt/m)

e^(kt/m) v = gm/k e^(kt/m) + C₁

v = C₁ e^(-kt/m) + gm/k

V₀ = C₁ + gm/k ....... ⇒ C₁ = V₀ - gm/k

v = (V₀ - gm/k) e^(-kt/m) + gm/k

v(20) = (0+32.174(160)/0.5))*e^(-0.5*20/160) - 32.174*160/0.5



Answer: v(20) ≈ 623.8 ft/s



x = ∫ v dt

= (V₀ - gm/k) ∫ e^(-kt/m) dt + gm/k ∫ dt

= (g(m/k)² - V₀ m/k) e^(-kt/m) + gm/k t + C₁

X₀ = (g(m/k)² - V₀ m/k) + C₁ .... ⇒ C₁ = X₀ - (g(m/k)² - V₀ m/k)

x = (g(m/k)² - V₀ m/k) (e^(-kt/m) - 1) + gm/k t + X₀

x(20) = (32.174(160/0.5)²)(e^(-0.5(20)/160) - 1) + 32.174(160)(20)/0.5



Answer: x(20) = 6302.8 ft



lim (t → ∞) (V₀ - gm/k) e^(-kt/m) + gm/k

= gm/k

= 32.174(160)/10

= 514.8 ft/s



Answer: terminal velocity = 514.8 ft/s



x = (g(m/k)² - V₀ m/k) (e^(-kt/m) - 1) + gm/k t + X₀

15000 = (32.174(160/10)² - (623.8)(160)/10) (e^(-10t/160) - 1) + 32.174(160)/10 t + 6302.8

t ≈ 14.85

⇒ reach the ground at 20 + 14.85 = 34.85 s



Answer: ≈ 34.85 s

Answer 2

there is quite a few methods. you will have right here situations, often: Seperable perfect Bernoulli Linear (First Order, 2d Order, and so on...) The observe "Order" skill what's the biggest spinoff in the equation? First, 2d spinoff? Linear skill not one of the words regarding the based variable are squared or taken to a pair different exponent. Homogenous Diff EQ's have a collection of stuff which includes derivatives on the left hand area and nil on the ultimate suited area. Inhomogeneous Diff EQ's have some thing different than 0 on the ultimate suited hand area. the final answer of a differential equation is the SUM of its homogenous answer and its specific answer. to discover the specific answer, there are a number of diverse methods for first and 2d order Diff EQ's (and you will cut back larger-order Diff EQ's to 2d order by applying substitutiton). besides the undeniable fact that, the least confusing thank you to sparkling up Diff EQ's consists of taking a Laplace remodel of that equation. seem up Laplace remodel. yet you should quite take a direction, it incredibly is a semester's properly worth of cloth to evaluate.

Answer 3

just a question to the answer .. why did the weight became the mass ? when w=mg : m=w/g ...