Distance traveled of a ball dropped in water.

[Smusko]

Homework Statement

A steel ball is released at the surface of the ocean and it takes 64 minutes for it to hit the bottom. The balls downward acceleration is a=0.9g-cv where g=9.82 m/s² and c = 3.02s^-1 and v is the speed. What is the depth of the ocean where the ball was released?

Homework Equations

The Attempt at a Solution

The ball will reach a top speed when v=(0.9g)/c

I am having trouble finding an expression for v with respect to time while the ball is accelerating since a depends on v.
When I am integrating a I don't know what to do with the -cv term to get an expression that depends on time

 

[tiny-tim]

Hi Smusko!

The balls downward acceleration is a=0.9g-cv
…
I am having trouble finding an expression for v with respect to time while the ball is accelerating since a depends on v.

dv/dt = 0.9g - cv …

where's the difficulty? ​

 

[Smusko]

Hi Smusko! dv/dt = 0.9g - cv …

where's the difficulty? ​

When you integrate that, what happes with v? v=0.9gt - cs? since ds/dt = v. It feels so wrong.
Or do you treat v as a constant and v=0.9gt - cst? That also feels wrong.

 

[tiny-tim]

When you integrate that, what happes with v? v=0.9gt - cs? since ds/dt = v. It feels so wrong.
Or do you treat v as a constant and v=0.9gt - cst? That also feels wrong.

uh??

separation of variables …

dv/(0.9g - cv) = dt​

 

[Smusko]

uh??

separation of variables …

dv/(0.9g - cv) = dt​

I am sooooo stupid it's almost depressing.
Thank you.