Calculating Terminal Velocity for a Ping Pong Ball: A Simple Equation

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To calculate the terminal velocity of a ping pong ball, the drag force equation can be used, incorporating the drag coefficient (approximately 0.44), air density, diameter of the ball, and gravitational force. The terminal velocity is reached when the drag force equals the weight of the ball, allowing for a straightforward calculation. An experimental method involves dropping the ball from a height and timing its fall to estimate terminal velocity, though this requires a significant drop height for accuracy. Alternatively, using known values for drag coefficient, diameter, gravity, and air density can yield a more precise calculation without the need for extensive experimentation. Overall, both theoretical calculations and practical experiments can effectively determine the terminal velocity of a ping pong ball.
profaith
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hey how do you find terminal velocity? let say if you need to find the terminal veolcity of a ping pong ball? anyone has any ideas? what kind of experiment can i conduct?
 
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profaith said:
hey how do you find terminal velocity? let say if you need to find the terminal veolcity of a ping pong ball? anyone has any ideas? what kind of experiment can i conduct?
The terminal velocity of spheres like ping pong balls are fairly easy to compute and measure. The motion of such a sphere in still air at standard atmospheric temp & pressure will have a Drag Coefficient approx constant at C_{drag} = (0.44). The force F_{drag} due to aerodynamic drag ("air resistance") when the sphere falls thru air under those conditions is given by:

:(1): \ \ \ \ F_{drag} \ = \ C_{drag} \, \rho_{air} \,<br /> \pi \, D^{2} \, V^{2} /8

where "ρair" is the air density, "D" the sphere diameter, and "V" its fall velocity thru still air. The sphere will rapidly reach terminal velocity at the "VT" such that Drag Force exactly balances gravitational force on the sphere, namely its weight "mg":

:(2): \ \ \ \ F_{drag} \ = \ m \, g

:(3) \ \ \ \ \Longrightarrow C_{drag} \, \rho_{air} \, \pi \, D^{2} \, V_{T}^{2} /8 \ = \ m \, g

You can easily solve for terminal velocity "VT" in the above equation.

The above value can be checked experimentally by dropping the sphere from an elevated level (tall ladder might do) in still air. Time the fall and divide the distance fallen (e.g., height of the level) by the time interval to determine approx terminal velocity. You'll need to determine experimentally (& with calculations) the drop height needed to obtain reasonably accurate results.


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xanthym said:
The above value can be checked experimentally by dropping the sphere from a tall ladder in still air. Time the fall and divide the distance fallen (e.g., height of the ladder) by the time interval to determine approx terminal velocity.

I presume that you're assuming V_T &lt;&lt; \sqrt{2gh} for that calculation, right? A web search gave a terminal velocity of about 10~ m/s for a ping-pong ball, so your ladder would have to be a good bit greater than 5 meters. Not crazy, but a bit of a stretch for everyday purposes.
 
wow this looks quite complicated. but thanks loads anyway! is there any other method/experiments tt we can use to determine the terminal velocity of the ping pong ball?
 
It's really not that complicated of an equation. We gave you C, your coefficient of drag as 0.44. D the diameter can be easily measured or even just looked up online (simply just the diameter of a ping pong ball). Gravity is known, mass can be measured, and the density of air can be found in charts for given air temperatures. Then, just solve for V. It would be much easier than measuring experimentally, and given the margin for error in the experiment, would probably be more accurate.
 
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 .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

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