if you dropped two objects with different weights from the top of the empire state building will they land at the same time?
if you dropped two objects with different weights from the top of the empire state building will they land at the same time?
They fall at the same speed.
Paul Osborn wrote:
They fall at the same speed.
Correct, but only if you ignore air resistance.
Terminal velocity accounts for air resistance. The two objects would accelerate at the same rate(assuming they have the same geometry). However the heavier object would stop accelerating later thus would have a higher terminal velocity.
You have two objects--same weight (sort of like mass)--different sizes. Size being sort of like geometic shape.
Say, like a five-pound tennis ball and a five-pound basketball.
Basketball has a lot higher drag (friction) coefficient, i.e., it will fall (accelerate) slower.
Tennis ball has a lower drag (friction against air) coefficient, i.e., it will fall (accelerate) faster.
Neither object will stop accelerating until they hit the ground, floor, roof of car ...
The heavier (denser) object will accelerate more rapidly.
Look up Stokes Law.
with the Millikan correction of course.
Two objects with the same air resistance and different masses? The more massive object falls faster.
ma = mg - F_friction
or
a = g - F_friction/m
So if m is large, a is large.
rgtegfd wrote:
Two objects with the same air resistance and different masses? The more massive object falls faster.
ma = mg - F_friction
or
a = g - F_friction/m
So if m is large, a is large.
Man, I'm an idiot, that's what I get for posting at 2AM...
You're looking for terminal velocity, not accelleration. Sorry 'bout that.
Watch out, y'all, I'm a physicist.
The really short answer to your question is yes, terminal velocity is directly proportional to the weight (mass * acceleration due to gravity). A longer answer would inform you that it is also inversely propotional to both the density of the fluid the object is falling through, and the cross-sectional area of the object.
The actual functional form depends on what kind of drag force you're assuming, i.e. proportional to velocity, velocity squared, etc. Typically, for many objects falling through air, you can assume a velocity squared drag force. The result is
like i said, Stokes law with the Millikan correction.