Drag forces on someone diving into a pool

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SUMMARY

The discussion focuses on calculating the time it takes for a diver to reach 2% of their original speed (0.14 m/s) after entering the water at 7.0 m/s, with a drag force defined as FD = (-1.00×10^4 kg/s)v. The diver's mass is 75 kg, and the forces acting on the diver are analyzed using Newton's second law, ΣF = ma. The participants clarify that the drag force coefficient is correct, despite initial doubts about its magnitude.

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  • Understanding of Newton's second law (ΣF = ma)
  • Familiarity with drag force equations
  • Basic knowledge of kinematic equations
  • Concept of variable acceleration due to drag
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  • Learn about differential equations in motion analysis
  • Explore numerical methods for solving motion problems with variable acceleration
  • Investigate the physics of fluid dynamics related to drag forces
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Homework Statement



You dive straight down into a pool of water. You hit the water with a speed of 7.0 m/s, and your mass is 75 kg.

Assuming a drag force of the form FD= (−1.00×104 kg/s)v, how long does it take you to reach 2% of your original speed? (Ignore any effects of buoyancy.)

Homework Equations



[tex]\Sigma[/tex]F=ma
V=Vo+at (?)

The Attempt at a Solution



I summed the forces in the y-direction (down is positive):
[tex]\Sigma[/tex]F=mg-FD=may

2% of 7 m/s = 0.14 m/s

But I don't know what to do from here. How do I solve for ay when the drag force varies with velocity? If ay varies, then how can I use the kinematic equations for constant a?

I figured this out the hard way after plugging in 7.0 m/s for v and solving for a, which is an absurdly large number. (-923 m/s2)
 
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Are you sure that FD=(-1.00x104)v?
This coefficient seems kind of large to be for drag, recheck the numbers.
 
Last edited:
Lazyshot,

I should've placed the negative sign outside the parenthesis, but yes, according to my book and MasteringPhysics, that is the coefficient.

FD=-(1.00x104)v
 

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