Sky Diver Free Fall: Speed, Distance, Limiting Velocity & Time Calculation

  • Context: Undergrad 
  • Thread starter Thread starter footballxpaul
  • Start date Start date
  • Tags Tags
    Modeling Stuck
Click For Summary
SUMMARY

The discussion focuses on the physics of a sky diver weighing 180 lb who falls from an altitude of 5000 ft and opens their parachute after 10 seconds. The equations of motion are derived considering air resistance, modeled as 0.75(v) before the parachute opens and 12(v) after. Key calculations include determining the speed at parachute deployment, the distance fallen, the limiting velocity after the parachute opens, and the total time in the air post-deployment. The correct setup involves separating variables in the differential equations derived from Newton's second law, leading to an exponential solution for velocity.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Basic knowledge of differential equations
  • Familiarity with concepts of weight and mass
  • Knowledge of air resistance modeling in physics
NEXT STEPS
  • Study the derivation of differential equations in physics contexts
  • Learn about air resistance and its impact on falling objects
  • Explore the concept of limiting velocity in fluid dynamics
  • Investigate numerical methods for solving differential equations
USEFUL FOR

Physics students, educators, and anyone interested in the dynamics of free fall and parachuting. This discussion is particularly beneficial for those studying mechanics and differential equations.

footballxpaul
Messages
30
Reaction score
0
A sky diver weighing 180 lb falls vertically downward from an altitude of 5000ft and opens the parachute after 10 s of free fall. assume that the force of air resistance is 0.75(v) when the parachute is closed and 12(v) when the parachute is open, where the velocity v is measured in ft/s.

a) find speed of the sky diver when the parachute opens
b) find the distance fallen before the parachute opens
c)What is the limiting velocity V_L after the parachute opens?
d) Determine how long the sky diver is in the air after the parachute opens

I have tried to set it up like -

(dv/dt)= (.75(v))/180-a and seperating and solving but answer doesn't make sense

and

V=32(t)-(.75v/180)t that didnt make sense either

I know I am probably setting this up wrong, I was just trying to do F=ma for the set up? please point me in the right direction. Thanks!
 
Physics news on Phys.org
I have no idea where you got that first equation- it has "180- a" and there is no reference to "a" in what you write before that. The second equation is almost right but you seem to be taking the positive direction of velocity downward which seems peculiar to me. More importantly, you have the acceleration on the right side equal to V rather than dV/dt. Finally, 180 pounds is weight not mass. Weight is the force due to gravity on a given mass: 180= mg so m= 180/g.

dv/dt= acceleration= force/mass and there are two forces acting here: gravitational force -mg and air resistance -(.75)v (- because the resistance is always directed opposite to the velocity it is resisting.)

m dv/dt= -mg- (.75)v so dv/dt= -g- (.75)v/m= -g- (.75)gv/180= -g(1+ (.74/180)v).

That is an easily separable equation. Solve for v(t) and determine v(10), the speed when the parachute opens. After the parachute opens the equation is
dv/dt= -g(1+ (12/180)v). Solve that using the previous v(10) as the initial value.

Hopefully, you will find that v(t) is an exponential with negative exponent. The "limiting velocity" is the limit as t goes to infinity. Even simpler, at the "limiting velocity", the velocity does not change so dv/dt= -g(1+ (12/180)v)= 0.
 
I was using a to be acceleration or dv/dt. But thankyou I knew I wasnt seeing it right.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
33K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
10K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
2
Views
7K
  • · Replies 8 ·
Replies
8
Views
21K
  • · Replies 2 ·
Replies
2
Views
2K