Free fall
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From: Don1 on 14 Sep 2005 20:39 Don1 writes: The formula for finding the velocity of free fall is: v = 2s/t; where s is the distance a body falls, and t is the time during which it falls. The formula for finding the distance a body free falls is d=vi+(a/2)t^2: Let's look at the progress of a body falling freely from rest (vi=0) after 4 successive intervals of time: During second 1, starting from rest, an object falls at 32'/sec^2; so that the distance it falls is d=vi + (a/2)t^2=16 ft. and it will attain a velocity of 2s/t=32 ft/sec. During second 2, an object falls at 32'/sec^2; so that the distance it falls is d=vi + (a/2)t^2= 64 ft. and it will attain a velocity of 2s/t=2x64 ft/2 sec = 64ft/sec. During second 3, an object falls at 32'/sec^2; so that the distance it falls is d=vi + (a/2)t^2= 144 ft. and it will attain a velocity of 2s/t=2x144 ft/3 sec = 96 ft/sec. During second 4, an object falls at 32'/sec^2; so that the distance it falls is d=vi + (a/2)t^2= 256 ft. and it will attain a velocity of 2s/t=2x 256 ft/4 sec = 128 ft/sec. Using these simple formulas it is possible to find the velocity and distance a body free falls during any period of time. Don
From: Randy Poe on 14 Sep 2005 20:49 Don1 wrote: > Don1 writes: > > The formula for finding the velocity of free fall is: v = 2s/t; where s > is the distance a body falls, and t is the time during which it falls. > The formula for finding the distance a body free falls is > d=vi+(a/2)t^2: Let's look at the progress of a body falling freely from > rest (vi=0) after 4 successive intervals of time: > > During second 1, starting from rest, an object falls at 32'/sec^2; so > that the distance it falls is d=vi + (a/2)t^2=16 ft. if vi = 0, not otherwise. > and it will > attain a velocity of 2s/t=32 ft/sec. > > During second 2, an object falls at 32'/sec^2; so that the distance it > falls is d=vi + (a/2)t^2= 64 ft. and it will attain a velocity of > 2s/t=2x64 ft/2 sec = 64ft/sec. 64 feet is the total distance it has fallen since t = 0. The distance it traveled in second 2 was 64 - 16 = 48 feet. > During second 3, an object falls at 32'/sec^2; so that the distance it > falls is d=vi + (a/2)t^2= 144 ft. Again, that's the total distance it has fallen (if vi=0). The distance just during second 3 is 144 - 64 = 80 feet. > and it will attain a velocity of > 2s/t=2x144 ft/3 sec = 96 ft/sec. > > During second 4, an object falls at 32'/sec^2; so that the distance it > falls is d=vi + (a/2)t^2= 256 ft. and it will attain a velocity of > 2s/t=2x 256 ft/4 sec = 128 ft/sec. > > Using these simple formulas it is possible to find the velocity and > distance a body free falls during any period of time. Yes, if correctly applied. This is the most correct post I've seen from you but there is still a slight misinterpretation of the meaning of the formulas. - Randy
From: odin on 14 Sep 2005 21:17 > The formula for finding the velocity of free fall is: v = 2s/t; where s > is the distance a body falls, and t is the time during which it falls. Is the moon in free fall? If yes, how does that work with your equation above. If not, then what the heck is it doing up there? > The formula for finding the distance a body free falls is > d=vi+(a/2)t^2: Let's look at the progress of a body falling freely from > rest (vi=0) after 4 successive intervals of time: You do not take into account the vector nature of the problem. What if velocity and force are not parallel? You do not take into account the fact that mass is not constant, but depends on velocity (relativistic). You do not take into account the fact that the force is not constant, but depends on displacement (universal gravitation) Also, even given all these simplifying assumptions, you have an error... you have: d = vi + (a/2)*t^2 you should have: d = vi*t + (a/2)*t^2 where d is the distance measured from when t=0. Not the distance measured for a given time interval as you describe...
From: Don1 on 14 Sep 2005 22:20 odin wrote: > > The formula for finding the velocity of free fall is: v = 2s/t; where s > > is the distance a body falls, and t is the time during which it falls. > > Is the moon in free fall? If yes, how does that work with your equation > above. If not, then what the heck is it doing up there? > It's falling, but missing, and passing by: Only to fall back again, and again. > > The formula for finding the distance a body free falls is > > d=vi+(a/2)t^2: Let's look at the progress of a body falling freely from > > rest (vi=0) after 4 successive intervals of time: > > You do not take into account the vector nature of the problem. What if > velocity and force are not parallel? Velocity is relative, and the force is centripetal. > You do not take into account the fact that mass is not constant, but depends > on velocity (relativistic). Says you. > You do not take into account the fact that the force is not constant, but > depends on displacement (universal gravitation) > The gravitation is in the direction of the free fall, and is its cause. > Also, even given all these simplifying assumptions, you have an error... > you have: > d = vi + (a/2)*t^2 > you should have: > d = vi*t + (a/2)*t^2 No vi is the point of beginning; the rate of the initial _inertial_ displacement l: It is equal to l/t; for vi = 0, it is 0/t. > where d is the distance measured from when t=0. Not the distance measured > for a given time interval as you describe... Yes, d= the total distance from wherever it starts. Don
From: Sam Wormley on 14 Sep 2005 22:52
Don1 wrote: > Don1 writes: > > The formula for finding the velocity of free fall is.... No you have to start with the differential equation F = ma, and this time acceleration is not constant.... it is quite a complicated calculation. Newton gave us all the tools we need. |