1. Feb 24, 2006

### entrepreneur787

The referee throws a jump ball vertically in the air. Besides the force
of gravity, the drag force acts on the ball. The drag force is given by
the product of the friction coefficient with the square of the velocity.
The friction coefficient is 0.1. The NBA ball mass is 0.6 kilograms. The
initial ball velocity is 10 meters/second. How much time does it take
for the ball to slow down to 2 meters/second?

So do you set
Fdrag = Fgravity

So far I have Fdrag = u x v^2
where Vi=10 and Vf=2 a= 9.8

Thanks

2. Feb 24, 2006

### quasar987

This is more mathematics than physics. Set up the equation of motion (F=mdv/dt), solve it. Find the particular solution by plugging your inital conditions. Set v = 2. Solve for t. You'll get two answers; trash the larger t.

The only physics involved is knowing why you trash the larger t.

3. Feb 26, 2006

### entrepreneur787

still don't understand

4. Feb 26, 2006

### siddharth

What part don't you understand? Can you post and show the steps you have taken till now?

5. Feb 26, 2006

### entrepreneur787

m*dv/dt=u*v^2
dv/v^2=u/m*dt
-1/v=u/m*t+C

so i set t=0 and Vi=10 and solve for C then plug in Vf and solve for t?

C=-1/10 but according to the second thread i'm suppose to get 2 t's.

Last edited: Feb 26, 2006
6. Feb 26, 2006

### quasar987

First of all, the drag force acts in the opposite direction of velocity. So since the ball is thrown up, and it is natural to take the y axis to point upward, you'll need to take $F_{drag} = -uv^2$.

Secondly, you're forgetting to account for the gravitational force in your equation of motion.

7. Feb 26, 2006

### entrepreneur787

doesn't m*dv/dt that's gravity

or

m*dv/dt=m*9.8-u*v^2

is so how can you isolate the v's on oneside?

8. Feb 26, 2006

you should write it as Vs on one side on the equation and ts on the other side so that you can integrate it