Calculating Maximum Height of Object Thrown Upward

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The discussion focuses on calculating the maximum height of an object thrown upward, specifically a mass M with an initial velocity Z, under the influence of gravity g and air resistance A. The correct differential equation governing the motion is m(dv/dt) = -mg - Av, where A represents air resistance as a force proportional to velocity. To find the maximum height, one must integrate the velocity function v(t) derived from this equation, applying the initial condition v(0) = Z. The participants clarify the importance of defining air resistance accurately to ensure correct calculations.

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I don't know much about classical physics, if I throw a rock or mass M upward with an initial velocity Z, with gravity g and air resistance A acting against it, is this equation the right one ;

[math]m\frac{dv}{dt} = (Z-g)m -Av[/math]

Then if I want to know the maximal height, I just need to integrate v(t) to find the distance at time t. Right ?

I'm really not sure about this equation, mostly because of the (Z-g) term. After all, it's about INITIAL velocity.
 
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You have to define what you mean by air resistance 'A', is this a force or a drag coefficient, is it a function of V or a constant?
Generally the equation you want is s = ut + 1/2 gt^2

See the formulae sticky thread for more equations.
 
If you define air resistance to be a force proportional to the velocity

Then the forces on it as it's going up are mg down and Av down, so you get

[tex]\frac{dv}{dt} = -g - \frac{Av}{m}[/tex]

You then have to solve this differential equation for v(t), using v(0) = Z as a boundary condition. Then you have all the information you need about the rock's movement.
 

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