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Ideally 2 objects of different mass and weight will hit the floor at same time because they face same gravitational pull and accelerate. Will the terminal velocity of both objects be same also? and with that theory does terminal velocity of every object is same ideally?

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Ah, a chance to get all sciency. Drop a feather and a tennis ball side by side and draw your own conclusion. –  dmckee May 22 '13 at 21:10
@Qmechanic I first though why would ZackAshton ask about terminal velocity after having read the Wikipedia article which basically contains the answer, then I saw you edited it in... –  Tobias Kienzler May 23 '13 at 8:27
This question was flagged by an anonymous user for being too localized. Generally, the questioner is expected to have performed a minimum Internet search before asking, such as, e.g., browsing the relevant Wikipedia pages for, say, "free fall" or "terminal velocity". See also this meta post. –  Qmechanic May 23 '13 at 9:03
@Qmechanic Good point. In that case en.wikipedia.org/wiki/… already answers the question... –  Tobias Kienzler May 23 '13 at 12:53
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4 Answers 4

Terminal velocity is reached when the drag force due to moving through air is equal (but opposite) to the gravitational force. Now, the gravitational force is proportional to the mass, while the drag force has nothing to do with mass, but everything to do with how large and "streamlined" the object is. Suppose object A is twice as heavy as object B. If object A also experiences twice the drag force as object B (at a given speed), then their terminal velocities will be the same.

To put it another way, let's suppose that the two objects have the same masses, and therefore the same weights; they have the same gravitational forces. The question becomes: do they have the same drag force?

Drag comes from the resistance of the air to an object's movement, so – all else being equal – something that's more streamlined will have less resistance. If one of this is shaped like a bullet, and one is shaped like a big hollow ball, the big ball will have the same amount of drag at low speeds as the bullet does at high speeds. So the ball's terminal velocity will be a lot lower.

(I've simplified a little; the force of buoyancy should be added to the drag force. But usually this is relatively small, so we can ignore it just for simplicity.)

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Only in a vacuum and in a constant/uniform gravitational field.

Air resistance affects terminal velocity within Earth's atmosphere. terminal velocity is when frictional forces balance gravitational forces.

Without some frictional force such as air resistance there isn't a terminal velocity in the normally accepted sense (there is a final velocity, e.g. when objects falling in a vacuum hit the ground)

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The relevant part is vacuum; the gravitational field does not need be constant/uniform (see my answer's footnote) - as long as both objects are at approximately the same position –  Tobias Kienzler May 23 '13 at 8:25
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Two objects of different mass only fall at the same rate in a vacuum. In the atmosphere, drag forces act on the object as it moves though the fluid (air). As velocity increases, these drag forces become larger. Terminal velocity is the point at which the drag force equals the force of gravity. Terminal velocity will depend on the mass, cross sectional area, and drag coefficient of the object as well as the density of the fluid through which the object is falling and gravitational accelleration.

To answer your question: Generally no. The differences in mass and other object properties will likely result in different terminal velocities. The best way to know is to do the math.

Here is a wikipedia article on terminal velocity.

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In extension to the other answers, I'd like to add some formulae:

Remember that an object's (of mass $m$) acceleration $a$ is given by the sum of all forces acting on it, $$m\cdot\vec a = \sum\vec F.$$

Terminal velocity means that $a=0$. When you drop something, the forces acting on it are usually the constant$^\dagger$ gravitational force $\vec F_g\approx m\cdot\vec g$ and the velocity $v$ dependent Drag. For "usual" objects in air and close to terminal velocity, it is the Newtonian drag $$\vec F_D = -\frac12\rho v^2C_DA\cdot \vec e_v,$$ pointing opposite to the velocity. Here $\rho$ is the density of air, $C_D$ the drag coefficient (which depends on the object's shape and orientation, e.g. for a sphere it's 0.47, while for a hollow hemisphere it lies between 0.38 (facing stream) and 1.42 (opposing stream)) and $A$ the cross-sectional area (perpendicular to the velocity) of the object. After a sufficiently long time the velocity is aligned parallel to gravity (since there is no other accelerating force), so equilibrium reads

$$m\cdot g \stackrel!= \frac12\rho v_\text{terminal}^2 C_D A \Rightarrow \boxed{v_\text{terminal} = \sqrt{\frac{2mg}{\rho C_D A}}},$$

i.e. the terminal velocity for two bodies at (approximately) the same position still differs if the term $C_D\cdot A/m$ is not equal.

Note that things severely change if e.g. an electric force is involved as in the Millikan experiment where one or more charged oil drops are first made floating due to an equilibrium of gravity and electrical force (to determine the ratio of their mass to their charge), and then the field is switched off and these drops accelerate to their terminal velocity in order to determine their mass and therewith the elementary charge $e$. Due to the drops being very small, the Newtonian drag is so negligibly small that here the Stokes drag $$\vec F_S = -6\pi R\eta\vec v$$ linear in velocity dominates ($\eta$ is air's viscosity, $R$ the drop's radius), yielding a different formula for the terminal velocity, $v = \frac{mg}{6\pi\eta R}$. And this is just one special case, in general things can become arbitrarily complicated...

$^\dagger$This is actually only an approximation for little change in height, the correct formula in the gravitational force of one heavy body (say, earth) is $\vec F_g = -m\cdot\frac{GM}{r^2}\vec e_r$ where $G$ is the gravitational constant and $r$ the distance from it's centre. While this doesn't change anything about terminal velocities being equal iff $C_DA/m$ is equal, that means that the terminal velocity actually depends on the distance to your local planet's surface.

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protected by Qmechanic May 23 '13 at 15:08

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