will two objects of different mass and size but of same shape reach the ground at the same time if dropped..?

Question

of same shape and same material..

which one would fall first??

Answer 1

If dropped in a vacuum, yes. Air currents and friction may slow the descent of a lighter or bigger object if dropped in an atmosphere. An experiment on the Moon using, IIRC, a hammer and a feather showed them falling at the same rate and hitting the ground at the same time.

EDIT: Think about it. If 2 objects of different materials, masses and shapes drop the same, 2 objects with the same material and shape should also. The material, mass and shape are irrelevant to the experiment since ALL objects fall at the same rate in a vacuum. All bets are off when it is done in air though.

EDIT2: To eyeonthescreen; Isn't that what I just said in my first 2 sentences?? LOL Thanks for the proof though. ;)

Answer 2

Not generally...here's why...

Net force on the falling bodies is f = W - D = M dv/dt (vdot); where W is weight = Mg and D is drag force = 1/2 rho Cd A V^2. Thus, vdot = (W - D)/M; where vdot = dv/dt is the acceleration in fall.

Depending on the height of release (h), two outcomes can happen. The object can reach terminal velocity where vdot = 0 = (W - D)/M; so that W = Mg = 1/2 rho Cd A V^2 = D. Or vdot <> 0 <> (W - D)/M; so that vdot = (W - D)/M > 0 (considering acceleration downward to be positive).

Let's consider the no terminal velocity case first. Here take the ratio Vdot/vdot = (W - D)/M//(w - d)/m; where the w's are weights for the large and small masses M and m, and the d's are the respective drag forces. From the SUVAT equation h = 1/2 vdot t^2; so that t = sqrt(2 vdot h), we see that the time to fall from a height h can be the same if and only if Vdot = vdot. This is clearly not the case when there is no terminal velocity, here's why.

From Vdot/vdot = (W - D)/M//(w - d)/m, we can write Vdot = vdot [(W - D)/M//(w - d)/m] = vdot (m/M)[(W - D)/(w -d)]. This means (m/M)[(W - D)/(w -d)] = 1.0; otherwise the acceleration Vdot <> vdot and the times to fall from the same height T <> t, for the big and small mass respectively. Without going through the math, we can show that in general the value will not be 1.0.

To start with, we assumed M > m; so that m/M < 1.0 at the outset. Thus, the only way for (m/M)[(W - D)/(w -d)] = 1.0 knowing that m/M < 1.0 is for the net forces ratio F/f to exactly equal M/m ALL THE TIME; so that (m/M)(F/f) = 1.0 = (m/M)(M/m). So does F/f = M/m all the time?

Nope...look here F/f = (W - D)/(w - d) = Mg - D/mg - d. This clearly shows that the only way F/f can = M/m all the time is for both D and d, the drag forces, to equal zero all the time. In other words, in a vacuum. And there you have it; so long as there is air drag on the falling objects, their times to fall will generally be different because the net forces on each one is different from the other.

What about when the objects reach terminal velocity and Vdot = vdot = 0 some time after release? When terminal velocity is reached, weight and drag forces are equal, but opposite. Thus, we have W = D and w = d.

The time to fall (T) with terminal velocity consists of the time to fall while still accelerating Ta and the time to fall when velocity V = constant, Tc. That is, T = Ta + Tc = sqrt(2 Vdot S) + (h - S)/V; where S = 1/2 Vdot Ta^2, which is the distance the object fell while accelerating at Vdot to reach termiinal velocity V. Note (h - S) is the fall distance remaining at constant terminal velocity V. We have similar equations for the smaller mass object.

Thus we can set T = Ta + Tc = sqrt(2 Vdot S) + (h - S)/V = sqrt(2 vdot s) + (h - s)/v = ta + tc = t; where the lower case variables are for the smaller mass m. I think you will agree that T <> t in general. For T = t to happen ALL THE TIME, and the two objects hit the turf in the same time, the sum of the two terms (acceleration and constant) on both sides would have to equal each other all the time.

For one thing, we've already found that Vdot <> vdot except in a vacuum. So, S/s = 1/2 Vdot Ta^2/1/2 vdot ta^2; and S = s(Vdot/vdot)(Ta/ta)^2 so that the only way S = s, when Vdot <> vdot, is for (Ta/ta)^2 = (vdot/Vdot) all the time. That's not likely to happen. Thus the acceleration terms in the time to fall equations for the large mass and for the small mass are unlikely to be equal. And since the second, constant velocity, term for each mass depends in part on S and s, those two terms are unlikely to be equal all the time either. So the only thing that could make T = t is for the two terms to exactly offset the differences in the two terms to keep T and t at the same value. Not likely to happen either.

The question you asked is a good one, which is why I took some time to answer it. The bottom line physics is this...in a vacuum, where there is no drag, the two objects would in fact hit ground level (h = 0), whatever that is, at the same time when release under the same force of gravity from the same height h. But with drag forces, the net forces (e.g., W - D) are most likely to differ and cause different acceleration rates (Vdot <> vdot) and different terminal velocities (V <> v) will result; so the times to drop will not usually be the same for the two objects with drag.

Answer 3

In general, no. A simple counterexample to the assertion that they would reach the ground at the same time is to consider a cigar dropped from the window of a Zeppelin. These clearly have different masses and sizes, but roughly the same shape. A Zeppelin is a rigid-body lighter-than-air aircraft kept aloft by buoyancy gained from displacing air with a rarer (less dense) gas such as helium or hydrogren. For the cigar gravity would overcome the buoyant force of the air, and the cigar would fall.

Answer 4

If both objects are solid, and made of the same material, then yes.

The non-obvious counter example is a small steel ball that is solid, will reach the ground sooner than a large steel ball that is hollow, and filled with Helium, since the such a large steel ball will tend to float.

Answer 5

no.since the mass varies,one which has more mass than the other,falls early