When we throw an object in the presence of gravitational attraction is the mass of the object proportional to the time of fall or not?

Question

I was confused when i saw two object of different masses (A and B)falling from the same height(h) but both of them strikes the ground at same time. Is this possible, that masses of the objects are independent from the time of fall? but how is this possible? if the Kinetic energy says that K.E=1/2mv^2 thus m is proportional KE, and and if the kinetic energy of two object is different then this would take different time, because kinetic energy is proportional to time.

Answer 1

... Is this possible, that masses of the objects are independent from the time of fall? but how is this possible?

It certainly seems possible, as you claim to have observed it!

if the Kinetic energy says that K.E=mgh thus m is proportional KE

It's not that $KE=mgh$ (in general), but rather $KE=mv^2/2$. If you consider the gravitational potential energy as well ($U=mgh$), then you can write down the total energy for either object as:

$$E=KE+U=mv^2/2+mgh$$

and if the kinetic energy of two object is different then this would take different time, because kinetic energy is proportional to time.

This seems to be the source of the inconsistency. There is no "kinetic energy is proportional to time" rule. If you roll a ball along a smooth surface, it will keep rolling with the same kinetic energy as time passes. A falling object will gain kinetic energy faster and faster as time progresses, in this case it will be quadratic, not (linearly) proportional.

To figure out what happens, you can use the fact that the energy is conserved ($\Delta E = 0$), and you will see: $$KE_{initial}+U_{initial}=KE_{final}+U_{final}$$

Now since the ball is not moving when you release it, $KE_{initial}=0$. Also, when it hits the ground $h=0$ so that $U_{final}=0$. We can specialize the above expression from conservation of energy to this case:

$$U_{initial}=KE_{final}$$ or $$mgh=mv^2/2$$

As mentioned in the comments, the mass appears on both sides here so we have finally that

$$gh=v^2/2$$

You will notice that I never said which falling object we were considering, and since the mass cancels out, this exact relationship must be true for both of the objects! From these considerations, you can find the velocity of an object after falling through an arbitrary height $h$, and if that velocity doesn't depend on the mass, then the whole time to fall can't depend on the mass either, as clearly two objects which always have the same velocity must both travel together to reach the ground at the same time.

Answer 2

It is true that if you neglect dissipative forces, all objects undergo the same acceleration while falling in a uniform gravitational field.It is because the gravitational force acting on an object of mass m due to another object of mass M equals GMm/r^2 when the are separated by a distance r.Let this force produce an acceleration a on m. Then, GMm/r^2=ma. Therefore a is independent of m. If two objects of different masses are dropped from the same height h, then , h=1/2*a*t^2,where t is the time of fall, which, as can be seen is same for any mass. And as for mechanical energy conservation, the heavier object had greater potential energy before it was dropped. It will also have greater kinetic energy because of it's greater mass, not any greater speed.