How Long Does It Take for a Basketball to Slow Down Due to Drag Force?

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In summary, the conversation discusses the physics behind a referee throwing a jump ball vertically in the air. It mentions the forces acting on the ball, including gravity and drag force, as well as the friction coefficient and the initial velocity of the ball. The conversation then turns to calculating the time it takes for the ball to slow down to a certain velocity. The solution involves setting up and solving an equation of motion, taking into account the direction of the drag force and the gravitational force. After solving, the final result is 0.52 seconds.
  • #1
entrepreneur787
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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

I'm really bad at physics can some Please help me

Thanks
 
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  • #2
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
still don't understand
 
  • #4
What part don't you understand? Can you post and show the steps you have taken till now?
 
  • #5
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.
 
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  • #6
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 [itex]F_{drag} = -uv^2[/itex].

Secondly, you're forgetting to account for the gravitational force in your equation of motion.
 
  • #7
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
first of all mg and the drag force is in the same direction so your Fnet=-mg-uv^2 = m dv/dt and then you need to get red of m by dividing both side with. Since you need to integrate the equation .
you should write it as V`s on one side on the equation and t`s on the other side so that you can integrate it
dV/-(9.8+uv^2/m)=dt and just integrate both sides ( v0 to V ) and (to=0 to t )
 
  • #9
thanks i got 0.52 secs..did anyone else who bothered to work out the problem in its entirety get that also
 

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