Okay, so there was this guy in my math class , asking my teacher a physics problem.
He wanted to know how long it would take an object to hit the ground, assuming it was from.. say... 10 meters , and had a weight. Our teacher said he didn't know since he hadn't done any physics for a while, so... I kinda got interested in it. (mostly because it will shut my classmate up if it gets answered, and I wont have to listen to him bugging about it 
)
I don't actually take physics at school, (yet), although I had an introductory course on it(just basics, but i have taken chemistry, and am trying to learn calculus on my own since my school wont let me take it hehe), so pardon any of my ignorance in advance.
I found this formula
d = Vi * t + (0.5)at^2
and from what I understood,
d = distance (m)
Vi = initial velocity (m/s)
t = time in seconds
a = acceleration (m/s^2) , which I also understood to always be 9.8 on earth.
Just as a note, I was using this http://www.physics247.com/physics-homework-help/free-fall.php which is where i got the formula from.
So, I decided, i would try to work out a problem mathematically, and then actually test it by dropping an object.
I measured a meter and dropped a pen from 1 Meter off the ground, and timed it, and did several trials, and got an average of 0.5s
So heres what i got when i worked it out.
1 = (0)t + (0.5)(9.
t^2
1 = 4.9t^2
0 = 4.9^2 - 1
At this point, i take what i have and solve using quadratic formula. ( But from what i read, theres another method you can use by just flipping the values to a negative, and moving it over.., which i'm assuming is the same thing, but involves less steps)
t = sqrt(-4(-1)(4.9)) / 2(4.9)
t = sqrt(19.6) / 9.8
t = 4.427 / 9.8
t = 0.451 s
So, i'm just wondering a few things,
1)Did I calculate that correctly?
2)Is mass a factor at all in this equation?
3)Is air resistance a factor at all in the equation?
4)If mass and air resistance are both factors, how would i work them into the equation, and how would i calculate the air resistance?
5)Also, is the area of an object a factor, i.e, a piece of unfolded paper, as opposed to a crumpled piece of paper. (Actually wouldn't that have to do with air resistance?)
Thanks in advance, also, sorry if anything i wrote is unclear, it's 3:10 am, and i'm kinda rusty on ANY physics knowledge i actually do have , but i'd like to brush up on that knowledge since i'm taking 2 physics course next year, and it'd be nice to have a bit of understanding going into the course. | [FuN]goku wrote: |
So, i'm just wondering a few things,
1)Did I calculate that correctly? |
Yep - that is correct according to Newton's laws and those tend to be what we use in this sort of problem (you could use Einstein but the 'relative' velocities are such that the answer would be the same to a first approximation.
| Quote: |
| 2)Is mass a factor at all in this equation? |
Nope - mass doesn't matter. Remember Galileo dropping 2 cannon balls of different mass off a tower - they both hit the ground at the same time.
The formula to calculate acceleration due to Gravity is given by Newton's famous expression
f=G*M1*M2/r^2
(f = force, G is gravitational constant, M1 is mass of object, M2 is mass of earth, r is distance between them)
Now, remember that f=m.a (force = mass * acceleration)
Therefore we get an expression for the force on the mass as:
M1*g = G*M1*M2/r^2
we can can cancel M1 on both sides to give:
g=G*M2/r^2 (gravitational acceleration = Gravitational constant multiplied by mass of earth, divided by the square of the distance [to the centre of the earth])
| Quote: |
3)Is air resistance a factor at all in the equation?
4..5.. |
No - this is the biggest inaccuracy here on Earth. Newton's laws are 'ideal' laws so they don't account for things such as wind, air resistance, cross sectional area etc. If you want to do 'real' calculations on bodies falling on earth then the maths gets VERY complicated.Okay! Thanks for clearing that up | Bikerman wrote: |
| [FuN]goku wrote: | So, i'm just wondering a few things,
1)Did I calculate that correctly? |
Yep - that is correct according to Newton's laws and those tend to be what we use in this sort of problem (you could use Einstein but the 'relative' velocities are such that the answer would be the same to a first approximation.
| Quote: | | 2)Is mass a factor at all in this equation? |
Nope - mass doesn't matter. Remember Galileo dropping 2 cannon balls of different mass off a tower - they both hit the ground at the same time.
The formula to calculate acceleration due to Gravity is given by Newton's famous expression
f=G*M1*M2/r^2
(f = force, G is gravitational constant, M1 is mass of object, M2 is mass of earth, r is distance between them)
Now, remember that f=m.a (force = mass * acceleration)
Therefore we get an expression for the force on the mass as:
M1*g = G*M1*M2/r^2
we can can cancel M1 on both sides to give:
g=G*M2/r^2 (gravitational acceleration = Gravitational constant multiplied by mass of earth, divided by the square of the distance [to the centre of the earth])
| Quote: | 3)Is air resistance a factor at all in the equation?
4..5.. |
No - this is the biggest inaccuracy here on Earth. Newton's laws are 'ideal' laws so they don't account for things such as wind, air resistance, cross sectional area etc. If you want to do 'real' calculations on bodies falling on earth then the maths gets VERY complicated. |
ömmm, air resistance is a important factor, but another point is the *density* of the falling body, because, if you drop a kilo of helium, you need a *lot* of patience.... | Bikerman wrote: |
Nope - mass doesn't matter. Remember Galileo dropping 2 cannon balls of different mass off a tower - they both hit the ground at the same time. |
I don't understand. If I would drop a ball of 10kg and then a ball of 20kg, wouldn't the one of 20kg hit the ground first?(1) The mass of the body is not a factor when you are calculating the time of fall. If the body is 2Kg or 200Kg they both fall on the ground at the same time. That is what Galileo proved by dropping two stones from the tower of Pisa.
(2) In your calculation the air resistance might have an effect if the body was lighter. Like feather will fall very slowly in air because it has more air resistance. But in normal case we neglect air resistance.
(3) The area of an object might increase or decrease air resistance. In general we don't account for the area.
You know Physics works on a lot of assumptions. It neglect small things which has a smalerl effect. Newton's laws are ideal and is calculated based on some assumptions.
| joostvane wrote: |
| Bikerman wrote: |
Nope - mass doesn't matter. Remember Galileo dropping 2 cannon balls of different mass off a tower - they both hit the ground at the same time. |
I don't understand. If I would drop a ball of 10kg and then a ball of 20kg, wouldn't the one of 20kg hit the ground first? |
actually, if you drop the 10kg ball first then it would hit the ground first 
think about it: if you drop a 10kg ball, and it´s slower than the 20kg ball, what happens, if you just stick both of them together? do they fall even faster because it´s a 30kg piece, or does the 10kg ball slow the other one down? the only way it can work out is, if both fall at the same speed.No-one ever believes that mass does not affect how fast an object falls (speaking ideally of course). This makes a great physics or even physical science demonstration. Take two objects of similar size but different masses (like bowling balls), set them up above electronic timing gates, release, and watch as amazement turns into understanding. That is one lesson they will not soon forget!
Have you ever taken a calculus class? It's always nice to know where those equations that you're using come from (although a little understanding in mathematics from calculus and beyond is useful... so hopefully you have that knowledge).
You first have to realize that acceleration is just a constant. Acceleration due to gravity is about -9.81 m/s^2. Note: For this problem, we will assume that acceleration due to gravity is always -9.81 m/s^2 (it does vary slightly) and that we are in a vacuum (calculating air resistance is a pain).
So you can now start out with simply acceleration:
acceleration = -9.81
What do you think you would get if you integrated that? Well one thing to realize is that the derivative of velocity is acceleration. Velocity is linear, with the slope being the acceleration. Therefore you get:
velocity = -9.81X + C.
C is an arbitrary constant of integration. But what is C (and X)? Well They're both relatively obvious. X will be time. This makes sense intuitively because if you're moving at an acceleration for X amount of time, you'll have a velocity of some magnitude. Well what if you started off with an initial velocity? Wouldn't you just add that to your velocity? You would. So C is initial velocity. You then get:
velocity = -9.81*time + initial velocity
You can also generalize these equations for other uses in basic physics to something like:
velocity = acceleration*time + initial velocity.
What would happen if you integrated this one more time? Well one thing that you need to realize (by either studying a lot of differentiation/basic calculus/basic physics or from intuition is that velocity is a derivative of a position. Does that makes sense? Your position will be dependent on the magnitude (or slope) of your velocity, right? So let's integrate it again.
position = -4.905X^2 + (initial velocity)X + C
What is C? Well first let's consider our position function. This is going to tell us the position of the object at time X. We're using this for objects being thrown up or down. This makes sense, right? The function is a parabola, which follows the format of objects being thrown. So what happens if the object being thrown up already has an initial height? Well that will be added to the position. So C will be initial height. Now you get the following:
position = -4.905X^2 + (initial velocity)X + initial height.
Now let's solve your problem. You're dropping an object from an initial height of 1 meters. I'm assuming that you're dropping the object and not throwing it down with an initial velocity, right? One more thing to realize is that the object will hit the ground when the position equals 0. So let's plug all of these into our equation.
0 = -4.905X^2 + 0X + 1
-1 = -4.905X^2
X^2 = 1/4.905
X = + or - sqrt(1/4.905)
X = + or - .452.
The - is extraneous so we're just left with .452 seconds.
P.S. To answer some questions posted. Mass is not a factor at all. Air resistance is a factor. If you drop two items of different masses but that have the same effect from air resistance, they'd fall at the same rate. Likewise, if you dropped a feather and a car in a perfect vacuum, they'd hit the ground at the same time. Gravity doesn't show favortism towards heavier objects!
Magnitude of Mass is not relevent. But air viscosity correction needed for precise result. For example: try calculating final velocity of an water drop that is formed 1Km above ground.
| chatrack wrote: |
| Magnitude of Mass is not relevent. But air viscosity correction needed for precise result. For example: try calculating final velocity of an water drop that is formed 1Km above ground. |
Well then you start getting into things such as terminal velocity...What about relativistic mass?
As the velocity increases, shouldn't the mass increase? And if the mass increases, shouldn't the gravitational force increase?
[Is this according to General Relativity or Special Relativity?]
| _AVG_ wrote: |
What about relativistic mass?
As the velocity increases, shouldn't the mass increase? And if the mass increases, shouldn't the gravitational force increase?
[Is this according to General Relativity or Special Relativity?] |
Well, we are talking General Relativity here (Special Relativity applies as an ideal case of GR with no gravity/mass).
The simple answer is yes - the kinetic energy of a body is added to the rest mass, and this gives a greater gravitational field (or, more correctly, a greater distortion of spacetime). | _AVG_ wrote: |
What about relativistic mass?
As the velocity increases, shouldn't the mass increase? And if the mass increases, shouldn't the gravitational force increase?
[Is this according to General Relativity or Special Relativity?] |
Here the question is falling of a body under gravity. The incress in mass is noticible
only if the speed is near that of light for an object. | Quote: |
What about relativistic mass?
As the velocity increases, shouldn't the mass increase? And if the mass increases, shouldn't the gravitational force increase?
[Is this according to General Relativity or Special Relativity?] |
No way yaar..its worth accounting only @ v.v.vhighh velocities...no way u account it here...
And Mass is not the factor...
| Quote: |
| I don't understand. If I would drop a ball of 10kg and then a ball of 20kg, wouldn't the one of 20kg hit the ground first? |
Earth pulls all objects with same accelration=g
..reason already explained above
But this isnt satisfactory...coz We have air here...
Air-a fluid..exerts a bouyancy force=Displaced vol*Density of air...
So here let density of air=p
here let density of ball=q
m=q*volume of ball*g
For ne ball net downlward force=
=[(m*g)-(p*Volume of ball*g)]
=Vol. of ball*g*[q-p]
net downward acceleration=
=F/mass
=Vol. of ball*g*[q-p]
--------------------
Vol. of ball*q*g
=same for both balls...
But in reality theres 1 more factor..Viscous resistance of air...That differs for both ballls...so in practical..1 ball will comer @ diff time depending on dimensions and density of balls..
