# Drag forces on someone diving into a pool

1. Sep 30, 2008

### Serik

1. The problem statement, all variables and given/known data

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.)

2. Relevant equations

$$\Sigma$$F=ma
V=Vo+at (?)

3. The attempt at a solution

I summed the forces in the y-direction (down is positive):
$$\Sigma$$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)

2. Sep 30, 2008

### Lazyshot

Are you sure that FD=(-1.00x104)v?
This coefficient seems kind of large to be for drag, recheck the numbers.

Last edited: Sep 30, 2008
3. Sep 30, 2008

### Serik

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