Calculating Terminal Velocity for a Falling Object with Air Resistance

In summary, at terminal velocity, an egg experiences a force due to air resistance of the form F = -kv. The only thing I'm not sure about is how to find the projected area of the object, but I think I can find that information using dimensional analysis. Thanks again!
  • #1
Arcarius
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Homework Statement


An egg of mass 0.10 kg is dropped from rest at a height h. It is known that the egg dropped experiences a force due to air resistance of the form F = -kv, where v is the velocity of the egg and k is a constant of proportionality equal to 0.01781 N*s/m. What is the terminal velocity of the egg?


Homework Equations



V = sqrt(2mg/pAC)
V_t is terminal velocity,
m is the mass of the falling object,
g is the acceleration due to gravity,
C_d is the drag coefficient,
\rho is the density of the fluid through which the object is falling, and
A is the projected area of the object.


The Attempt at a Solution


The only thing I know about terminal velocity is the formula mentioned above, but it seems to me that I'm missing information (I don't have the projected area of the object, the density of the air, etc.)
The only thing I can think of is that the air resistance covers all of these and can be substituted in, giving me V = sqrt(2mg/-kv). However, I checked this with dimensional analysis and it doesn't yield the correct units. This problem is really confusing me, so I'd appreciate any help on it!
 
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  • #2
Identify the forces acting on the egg as it falls. How do the forces relate to the acceleration of the egg? At terminal velocity, what is the acceleration?
 
  • #3
TSny said:
Identify the forces acting on the egg as it falls. How do the forces relate to the acceleration of the egg? At terminal velocity, what is the acceleration?
Thank you for the response!
The forces acting on it are kv going upwards and mg going downwards. I'm taking the downwards direction as positive, meaning that the net force acting on it is mg-kv which is equal to ma. So the acceleration is equal to (mg-kv)/m, and I know that it is 0 when it reaches terminal velocity. So, I got (mg-kv)/m = 0, and solving for v, I got mg/k.
Is this correct?
 
  • #4
That's correct.
 
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  • #5
TSny said:
That's correct.

Awesome, thank you for your help! Your questions really helped guide me :)
 

1. What is terminal velocity?

Terminal velocity is the maximum speed that an object can reach when falling through a fluid, such as air or water. It is when the force of gravity is balanced by the resistance of the fluid, resulting in a constant velocity.

2. How is terminal velocity calculated?

Terminal velocity can be calculated using the equation: Vt = √(2mg/ρAC), where Vt is terminal velocity, m is the mass of the object, g is the acceleration due to gravity, ρ is the density of the fluid, A is the cross-sectional area of the object, and C is the drag coefficient. This equation takes into account the forces of gravity and air resistance.

3. What factors affect terminal velocity?

The factors that affect terminal velocity include the mass and shape of the object, the density and viscosity of the fluid, and the effects of air resistance (such as friction and drag). These factors can vary depending on the specific situation and can affect the final terminal velocity of the object.

4. Can an object reach terminal velocity in a vacuum?

No, an object cannot reach terminal velocity in a vacuum because there is no surrounding fluid to create a resistance force. In a vacuum, an object will continue to accelerate due to the force of gravity until it reaches its maximum potential velocity.

5. How does terminal velocity affect skydivers?

Terminal velocity plays a crucial role in the sport of skydiving. Skydivers aim to reach terminal velocity during their freefall, which is typically around 120 mph. This allows them to maintain a stable and controlled descent before deploying their parachute. Understanding the concept of terminal velocity is important for skydivers to safely and accurately plan their jumps.

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