How Do You Calculate the Force of Air Friction on a Falling Basketball?

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To calculate the force of air friction on a falling basketball, first determine the gravitational force using the equation fg = ma, which yields 24.5 N for a 2.5 kg basketball. The actual acceleration of the ball is 9.4 m/s², which is less than the gravitational acceleration of 9.8 m/s² due to air resistance. The difference in acceleration indicates the presence of air friction, which can be calculated using the equation ma = mg + Fdrag. By rearranging this equation, Fdrag can be found as Fdrag = mg - ma. The presence of air causes the basketball to experience a net force that reduces its acceleration compared to free fall in a vacuum.
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NEWTON'S LAWS: please help!

:eek:

so I am kinda stuck.

problem:

a 2.5kg basketball is dropped from the top of a building. its acceleration is found to be 9.4m/s2 as it drops to the ground. What is the force of air friction on the ball as it falls?

so like...

a=9.4
m-2.5kg
ffs=?
fg=ma
=(2.5)(9.8)
fg= 24.5N

now how do i find Ffs?
 
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What would the acceleration be if there were no air (like in a vacuum)? What causes the difference? How can you use F=ma to solve for the air resistance force and acceleration?
 
from ma=mg+Fdrag that simple. a is given, g is 9.8
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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