Vertical Projectile with air friction

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Discussion Overview

The discussion revolves around determining the maximum height a ball can reach when thrown vertically upward while considering air friction. Participants explore the effects of different friction models on the motion of the ball, including both linear and quadratic drag forces.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests starting from the equation of motion, incorporating friction as an additional force term, leading to the equation F = ma = -mg - cv.
  • Another participant agrees with the equation of motion but seeks clarification on how to find the maximum height.
  • A later reply indicates that solving the differential equation is necessary to find the height.
  • One participant proposes a change of variables to facilitate solving the differential equation, integrating from 0 to the height h.
  • Another participant introduces the concept of turbulent air resistance, suggesting the equation m \ddot{x} = -mg - c \dot{x}^2 for high velocities.
  • There is a discussion about the conditions under which turbulent flow occurs, with one participant mentioning the Reynolds number and its implications for the drag model.
  • One participant notes that the drag force could be proportional to velocity for low speeds and to the square of velocity for high speeds, indicating a complex solution for maximum height.
  • Another participant suggests using integrals to solve the problem, specifying the time domain for the integration.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate model for air resistance, with some advocating for linear drag and others for quadratic drag. The discussion remains unresolved regarding the best approach to find the maximum height under these conditions.

Contextual Notes

Participants acknowledge the complexity of the problem, including the need to consider different drag models based on the velocity of the ball and the conditions for turbulent flow. There are unresolved mathematical steps related to solving the differential equations presented.

madking153
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hi,

If we throw a ball vertical upward, we can easily find the maximum height if ignore the friction..

So if we switch on the friction, how can we find the max height ?
 
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It will be just like the way you would go find the maximum height the ball would reach without the friction. That is, you will start from F = ma but now instead of F being only mg, there will have to be another force term describing the friction. Most likely, it will be some constant times the velocity of the object. So, your equation of the motion will be

F = ma = -mg-cv

Now is a math problem.
 
i had the same equation motion, but how we can find the max height ?
 
You find the height by solving the differential equation.
 
This is the differential equation you need to solve.

[tex]m \ddot {x} = mg - c \dot{x}[/tex]
 
I believe that the best way to solve this is to make the change of variables
[tex]\ddot{x} = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}[/tex]
and integrate from 0 to the height h then solve for h.
 
Integral said:
This is the differential equation you need to solve.
[tex]m \ddot {x} = mg - c \dot{x}[/tex]
actually for a ball in air, the air resistance is turbulent and the equation will be:[tex]m \ddot {x} = -mg - c \dot{x}^2[/tex]
 
krab said:
actually for a ball in air, the air resistance is turbulent and the equation will be:[tex]m \ddot {x} = -mg - c \dot{x}^2[/tex]
I was thinking along those lines also but was not sure enough to make the correction.
 
krab said:
actually for a ball in air, the air resistance is turbulent and the equation will be:[tex]m \ddot {x} = -mg - c \dot{x}^2[/tex]


What satisfies the 'turbulent' condition?
 
  • #10
If it is [tex]\dot{x}^2[/tex] , you would have to consider the horizontal velocity at each instant too.
 
  • #11
i will consider both conditions .. if the initial velocity is not too high - then is propotional to v , if not ( high velocity ) then is proportional to v^2...


i got a long a strange solution for max height...
 
  • #12
whozum said:
What satisfies the 'turbulent' condition?
It's turbulent if the Reynolds is greater than about 30. This happens at very low speeds, so you can safely ignore the laminar case.
 
  • #13
Answer.

You can use intergral to solve this problem. At this time, the domail is from t1 to t2.
 

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