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

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Homework Help Overview

The discussion revolves around the dynamics of a basketball thrown vertically, specifically examining the effects of drag force and gravity on its motion. Participants are exploring how to calculate the time it takes for the basketball to decelerate from an initial velocity of 10 meters/second to a final velocity of 2 meters/second, considering the drag force proportional to the square of the velocity.

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Approaches and Questions Raised

  • Participants discuss setting up the equation of motion using the drag force and gravitational force. There are attempts to derive the relationship between velocity and time, with some questioning the correct application of forces and integration methods.

Discussion Status

Some participants have provided guidance on setting up the equations and integrating them, while others express confusion about specific steps and the implications of the results. There is an acknowledgment of multiple interpretations regarding the integration process and the resulting time values.

Contextual Notes

There is mention of the drag force acting in opposition to the direction of motion, and participants are navigating the complexities of integrating the equations of motion while accounting for both drag and gravitational forces. Some constraints include the initial conditions and the need to discard certain solutions based on physical relevance.

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