Calculating Maximum Height of Object Thrown Upward

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The discussion centers on the physics of an object thrown upward, specifically the correct formulation of forces acting on it. The initial equation presented is questioned due to the inclusion of the term (Z-g), which may not accurately represent the scenario. Clarification is needed on the definition of air resistance 'A', whether it is a force or a drag coefficient. A more appropriate equation is suggested, incorporating gravity and air resistance as forces acting on the object. Solving the differential equation for velocity with the initial condition provides the necessary information to determine the object's motion and maximum height.
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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 coefficent, 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

\frac{dv}{dt} = -g - \frac{Av}{m}

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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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