Air Resistance and Terminal Velocity

  • Thread starter Thread starter Hazeel
  • Start date Start date
AI Thread Summary
The discussion centers on how the height from which an object is dropped affects its velocity, specifically through the equation v² = 2gh, which illustrates that an increase in height results in a greater velocity just before impact. Participants clarify that this equation derives from the conversion of potential energy to kinetic energy during free fall. Additionally, air resistance is mentioned as a factor that influences terminal velocity, which occurs when the force of air resistance equals the object's weight. The conversation highlights the importance of understanding both gravitational effects and air resistance in the context of falling objects. Overall, the relationship between height and velocity is established, while acknowledging the complexity added by air resistance.
Hazeel
Messages
3
Reaction score
0
We have been given an assignment with the question:

"How does the height of a body falling through the air effect its velocity?"

It's clear that as the height the object is dropped from, the greater the velocity of the object will be. However, I was wondering if anybody could give me any help with some equations/maths that proves this. I realize this is simple stuff, however I can't quite get my head around the thought and think of the equation that would be used.

I would appreciate any equation based help to be of a relatively simple nature as my Physics is not very advanced.

Also, any help with how the below equation can help to prove that height will increase the velocity of a falling object would be appreciated to.

V2 = 2gh

V2 being V squared.

Thank you,
 
Physics news on Phys.org
The equation you wrote comes from:

<br /> \frac{1}{2}m v^2 = m g h<br />

This equation represents the fact that potential energy (right hand side) is converted to kinetic energy (left hand side) on falling a distance h through the gravitational field. Cancelling m on both sides and multiplying by 2 gives what you wrote:

<br /> v^2 = 2 g h<br />

Which expresses the change in speed when falling a distance h.
 
Ah ok, that helps a lot. Thank you!
 
Can anyone else provide any other information please?
 
Hazeel said:
Can anyone else provide any other information please?

Maybe it could be stated a bit more precisely - v^2 = 2gh represents the speed the object has just before it touches the ground, and that very speed depends on the height h from which the object is dropped.

There is no other information, you got the answer to your question in the posts above. If you meant something else, maybe you should rebuild your question. :smile:
 
Hazeel said:
We have been given an assignment with the question:

"How does the height of a body falling through the air effect its velocity?"

It's clear that as the height the object is dropped from, the greater the velocity of the object will be. However, I was wondering if anybody could give me any help with some equations/maths that proves this. I realize this is simple stuff, however I can't quite get my head around the thought and think of the equation that would be used.

I would appreciate any equation based help to be of a relatively simple nature as my Physics is not very advanced.

Also, any help with how the below equation can help to prove that height will increase the velocity of a falling object would be appreciated to.

V2 = 2gh

V2 being V squared.

Thank you,
I'm not sure what topic you are studying, but it may be that this is a question about air resistance, not energy conservation. Skydivers achieve a terminal velocity when the force of air resistance becomes equal to their weight. Air resistance will vary with height as well as speed. Where do you think the resistance will be greatest? Why?
 
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...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top