The Parachute Problem (adapted from Douglas B. Meade’s “ODE Models for the Parachute Problem”) Noam Goldberg Craig Kaplan Tucker Riley Outline Introduction to the Parachute Problem Modeling the Descent Derivation of Equations Application of Equations Results The Parachute Problem Newton’s Second Law: F ma x x0 v Figure 1: Forces acting on a skydiver in flight Mathematical Model of Falling Object m x' ' m g kx' Or, in the case of the Parachute Problem: m g k x ' 1 m x" m g k 2 x ' t td t td Equivalent First-Order System k v ' g v, m v(0) 0, x ' v, x ( 0) x 0 , Solving for Velocity v' g k v m k k v' g v → v' v g m m k ↓ v' v g v' P(t )v Q(t ) m (We can use an integrating factor!) k P(t ) , m Q(t ) g Integrating Factor P(t ) dt e e v e k dt m 1 Q(t ) dt k t m ( g )e k t m ge k t e k t m k t m dt dt k t k t m e . k t m gm 0 Finding the Integrating Constant e gAe k k t (General m velocity mg v(t ) gAe k Using the initial condition: v(0) 0 We can solve for A: m A k . equation) (2.15 Velocity Equation Before Deployment mg e v(t ) k1 k1 t m 1, t td Solving for Position x (t ) v (t ) dt k u-substitution, Using our velocity equation, t mg m and e condition: the initial 1dt x(0k) x , 0 k we seethat mg m t 2 e 1dt. k m t m kg k x(t ) x0 2 t 1 e k m Solving for Velocity After Deployment At t=td we have a new initial condition: v(td ) v0 Plugging this value into the general velocity equation (the one we had before plugging in ICs), we obtain: k td mg v(td ) v0 gAe m k Solving for A, we get: A e k td m m v0 k g Solving for Velocity After Deployment Plugging A into the original velocity equation, we find that: k k 2 td t mg mg m2 td t v(t ) e em v0 k2 k2 To get v0, we plug t=td into our equation for velocity before deployment: k1 m g m td v 0 v(td ) 1 e k1 Solving for Velocity After Deployment k1 td mg mg mg e m 1 v(t ) e e k k2 k2 k2 k2 k 2 k1 td t m g 1 td m g m g m t d t m m m v(t ) e 1 k1 k 2↓ e k 2 e m gk2 m tkd 2 m t td m g k1 mt td e e e v(t ) 1 1 k k k1 k1 k 2 2 t t 2 t t t mg m g d d d e m 1e m e m v(t ) 1 k1 k 2 k2 t d t m k2 k td 2 t m m Velocity Equation for Whole Jump m g k1 t e m 1 k 1 v(t ) k1 k2 k2 t t t t t d m g m g d d m m m e 1e 1 k e k2 1 t td t td Applications of Derived Equations h = 3500m td = 60 seconds m k2 5 k1 1 g 9 .8 2 s m 3 m 6 1. When the chute is pulled, what is the elevation and velocity of the skydiver? 2. How long is the total jump? 3. At what velocity does the skydiver hit the ground? 1. Elevation and velocity at time of deployment 1 ( 60 ) 36 9.8 1 x(60) 3500 (60) 1 e 6 1 6 x60 325m 1 6 9.8 6 ( 60 ) v(60) e 1 1 v60 58.8 m/s 2. Total length of jump Setting x(tf) = 0, solve for tf 5 tf 9 9.8 5 x(t f ) 0 324.784 t f 1 e 3 25 3 t f 55.835 60 116 s 3. Final velocity Solve for v(116) 1 5 5 60 116 60 116 60 6 9.8 3 9.8 6 3 3 e e v(116) 1e 1 1 5 v(116) 5.88 m/s Graph of negative velocity versus time -velocity (m/s) time (s) Applications of Derived Equations For h = 3500m td = 60 seconds k1 1 m 6 k2 5 m 3 g 9 .8 m s2 x(td) = 325m, v(td) = -58.8 m/s ≈ 130 mi/hr 2. tf = 116 s 3. v(tf) = -5.88 m/s 1. Conclusion Verified that Meade’s equations are correct by deriving them ourselves Used derived equations to find various velocities and positions, and total time of a typical jump (Meade) FOR SALE: PARACHUTE ONLY USED ONCE NEVER OPENED SMALL STAIN Works Cited Blanchard, Paul, Robert L. Devaney, and Glen R. Hall. Differential Equations. Third Edition. Belmont, CA: Brooks/Cole, 2006. Print. Meade, Douglas B. “ODE Models for the Parachute Problem.” Siam Review 40.2 (1998): 327-332. Web. 27 Oct 2010.