Vertical Projectile with air friction

AI Thread Summary
To determine the maximum height of a ball thrown vertically upward with air friction, the motion equation must include a friction term, typically represented as a constant times the velocity. The equation of motion becomes m * d²x/dt² = -mg - cv for linear drag, or m * d²x/dt² = -mg - c * (dx/dt)² for turbulent drag. Solving for maximum height involves addressing the differential equation, often requiring a change of variables and integration. The turbulent condition is satisfied when the Reynolds number exceeds approximately 30, which occurs at low speeds. The discussion emphasizes the importance of correctly modeling air resistance to accurately calculate the maximum height.
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.

m \ddot {x} = mg - c \dot{x}
 
I believe that the best way to solve this is to make the change of variables
\ddot{x} = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}
and integrate from 0 to the height h then solve for h.
 
Integral said:
This is the differential equation you need to solve.
m \ddot {x} = mg - c \dot{x}
actually for a ball in air, the air resistance is turbulent and the equation will be:m \ddot {x} = -mg - c \dot{x}^2
 
krab said:
actually for a ball in air, the air resistance is turbulent and the equation will be:m \ddot {x} = -mg - c \dot{x}^2
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:m \ddot {x} = -mg - c \dot{x}^2


What satisfies the 'turbulent' condition?
 
  • #10
If it is \dot{x}^2 , 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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