laws of motion

Neither. It is all about about initial velocity minus the acceleration of gravity. The terms involving mass appear in both sides of the equation, and they cancel out, as Galileo figured out (but Archimedes didn't).

Not exactly if experiment occurs in air since drag depends on the form and the cross area of objects as well as the speed relative to environment.

If an object is bigger (same density but bigger mass) then assuming same form the cross area will be (M/m)^(2/3) bigger so that the drag will be different and it comes in the equation for the dynamic forces balance.

... but the OP said "suppose the air exerts a constant and equal force of resistance on the two bodies" - unrealistic, yes, but a pre-condition in this case.

Ok

The heavier object will go higher.

Before I tell you why, you must explain to me the inanity of air drag being independent of velocity. If you cannot, then I must conclude you are a student and looking for homework help.

The air drag of course depends on velocity, but at all times the velocities of these two objects are the same. The OP stipulated that the air drag is alike for each object, not for each velocity.

H = vt - (g/2) t²

If the mass of each object is somewhat proportionate to ite area, the simplifying assumption of like drag for each object won't be too unrealistic.

Well, no, if there is air drag, the velocities are not the same except at the beginning. And the OP stipulated a constant drag force. I can believe he didn't really mean that, but was merely quoting a homework problem, but the heights are greatly different when you include any drag.

i can't explain.

Forgive me for being blunt. Such a drag force is physically impossible. Perhaps you have stated the question incorrectly. Perhaps you have obtained information from another. In any event, I have given you the correct answer that the heavier object goes higher. Cheers.

If you ask for a solution fo such a problem you should be able to find the expalnation.

In fact there are several ways to do it.

One of those is based on an energy balance.

M Vo^2/2 = G*H + ∫Fr*dH which means tha initial kynetic energy= work in the garavitational fireld + work done against resistace Fr at stop height. → H= (M*Vo^2/2-∫Fr*dH)/G since G= M*g → H= Vo^2/(2*g)-(∫Fr*dH)/G

The second term is reduced at higher "G" thus H will be bigger.

But this is a NOT realistic case which indicates a home work most probabaly.

IF there is NO air drag...

Both objects will travel the same height as posted in #1.


IF "... the air exerts a constant and equal force of resistance on the two bodies"

Then the larger mass will go higher because a=F/M and the larger mass will experience a smaller acceleration due to the air.

OP... I hope this isn't homework. To clear up some of the ambiguity, try to re-write your problem as follows:

Two spheres of equal size, but different uniform densities, are launched vertically with the same initial subsonic velocity. Gravitational acceleration is 1g , air density rho is typical for sea level at 20 deg C, and force from air resistance is derived from the standard air drag equation {Fd=rho*Cd*Area*velocity^2} .

Which sphere goes higher?

Answer: In this case the higher mass sphere will still reach a greater height because the drag force has less affect on the larger mass.

height and velocity versus time (s) for an initial velocity of 100m/s and in the earth gravitational field with g=9.81 m/s^2.

The height for zero drag is the most important and the velocity is linear versus time. When body again on ground its velocity is equal to the one at start.

When a drag is considered 3 aspects do change:

- height decreases with mass decrease since the cross area decreases less rapidly as the mass (considered same geometry and density) so that the drag influence grows up.

- velocity is not any more linear versus time

- when body again on ground velocity is not any more equal to the start value, it is less.

Between the curves for a non zero drag it was considered that the masses change with a factor of 2.