I'm currently in an argument over this... if you throw a ball directly
up in the air (ignore air resistance) what forces are acting on the
ball?  More precisely, when the ball reaches the maximum height what
is the acceleration, I think it is still -9.8 m/sec^2 but my friend
says that it is 0, to me that is the same type of theory that if you
are passing a car you are going the same speed when you are next to
it, what I really need is some documentation either way as prove to
see who won the bet.

mathisfun...

Your friend is correct. At the maximum height, the NET force
operating on the ball is zero - that is, the combined forces
of both the force applied to the ball when it was thrown, and
the force of gravity, which you correctly identify as 9.8 m/sec^2,
are, at that moment, balanced out and cancel each other out.

This same principle is used to expose astronauts to a weightless
environment, by flying a large plane as close to straight up
as is possible, and then cresting an arc as it begins to fly
downward. At the top of the arc, the astronauts can experience
weightlessness.

The formula for this is: Fnet = m * a  where Fnet is net force,
m is mass, and a is acceleration. See this page on Newton's 
second law of motion in this Glenbrook H.S. Physics discussion:
http://www.glenbrook.k12.il.us/gbssci/phys/Class/newtlaws/u2l3a.html


When there is more than one force applied to the object, the
results are calculated by using vectors, which are like arrows
with direction and with the amount of force indicated by the
length of the arrows. If the force applied to the ball in
throwing it upward were maintained at a constant level, the
vectors would be of equal length, with one pointing up, and 
the other downward, and the object would exhibit the stationary
quality of inertia. This happens when you hold the ball, at 
the same height, in the palm your hand. The force you are 
applying to hold the ball up is equal to the force of gravity
pulling it down.

If you stop applying that force, your arm will fall, and the
ball will respond to the constant vector of gravity, and fall
to the ground.

When you throw the ball, your briefly apply a force which is
greater than gravity, but, since you cease applying it, the
upward acceleration of the ball is eventually overcome by the
constant acceleration of the force of gravity. At the uppermost
height, those forces are momentarily equal, and the NET force
applied to the ball at that moment is zero.

A discussion of vectors in the calculation of NET force appears
on the next page of the discussion cited above:
http://www.glenbrook.k12.il.us/gbssci/phys/Class/newtlaws/u2l2d.html#Practice

"Situation C" precisely represents the situation you're
describing, and if you click on the button to see the answer
it notes:

"The net force is zero Newtons. All the individual forces
 balance each other (i.e., cancel each other out)."

I hope you didn't lose any money... : )


Please do not rate this answer until you are satisfied that  
the answer cannot be improved upon by way of a dialog  
established through the "Request for Clarification" process. 
 
A user's guide on this topic is on skermit-ga's site, here: 
http://www.christopherwu.net/google_answers/answer_guide.html#how_clarify 
 
sublime1-ga


Additional information may be found from an exploration of
the links resulting from the Google searches outlined below.

Searches done, via Google:

"force equals"
://www.google.com/search?q=%22force+equals%22

---

Request for Answer Clarification by mathisfun-ga on 04 Oct 2005 18:53 PDT

I'm still having troubles, from my understading there is no force
acting upward on the ball after it leaves your hand... maybe if I put
some numbers in I will be able to better understand your explination. 
Say we have a 1 kg ball, now when throwing it up say we use a impulse
of 49 newton-seconds, and as soon as you let go of it it starts
slowing down at 9.8m/sec^2 meaning the net force is downward with
magnitude of 9.8 Newtons which implies the only acceleration is due to
gravity, and this force of 9.8 N stays constant for the 5 seconds for
the ball to reach the peak of it's path, and also on the return path
the Force is 9.8 N in the same direction, my question is what force is
added as an upward force against gravity for that point where the ball
is at its maximum, also where does it go for the ball to be able to
start moving again.

---

Clarification of Answer by sublime1-ga on 04 Oct 2005 21:46 PDT

mathisfun...

I apologize for missing a fine point in your question.
I will admit to having been rushing to complete the 
answer in order to make an appointment on time.

I focused on the forces acting on the ball, and neglected
your focus on the acceleration on the ball at maximum
height. francf-ga has provided a very good explanation of
why the acceleration of the ball is -9.8 m/sec^2 for the
entire time it is in the air, once it leaves your hand,
which makes YOU correct in your assertion.

Since francf-ga's comment was added after your Request
for Clarification, it's not clear if it satisfied your
need for further explanation. It certainly satisfied
mine.

If you'd prefer that I remove my answer, I can ask the
editors to do so. Likewise, if there's something that's
not clear in francf-ga's explanation, I can try to
clarify that.

Let me know...

sublime1-ga

---

Request for Answer Clarification by mathisfun-ga on 04 Oct 2005 22:02 PDT

I'm fine with giving you the payment and even a good rating, but if
you were paying attention to forces shouldn't and acceleration was
constant then the force acting on the ball would have been constant as
well?  If you get back to me with either agreeing or disagreeing with
this (and if you disagree explain what other forces are in play
besides the impulse while throwing the ball) I have no problem giving
the small fee of $5, after all you did put probably that much time in
it.

---

Request for Answer Clarification by mathisfun-ga on 04 Oct 2005 22:03 PDT

and as a side note, the funny thing is that it wasn't really a bet
between me and a friend but an argument between me and my physics
professor, I just didn't want to post that in case some would be
biased that the Dr. would naturally be correct

---

Clarification of Answer by sublime1-ga on 04 Oct 2005 22:35 PDT

mathisfun...

That's funny about it being between you and the prof!
And what you did to avoid bias makes perfect sense.

Yes, once the ball leaves your hand, having had a force
applied to it, the only force continuing to act on it
is the downward force of gravity, and, in that sense,
the ball appearing at rest at the top of the travel
is not the same as a ball at rest in your palm due
to a constant upward force opposing the constant 
downward force of gravity. 

The opposing vectors of force would only apply at 
the very beginning when a longer vector (more force
than that of gravity) was (momentarily) pointing up
in direct opposition to the downward (and unchanging)
vector of gravity. Once the ball leaves the hand, there
is really only one vector in the picture, which gradually
overcomes the effect of the brief, initial upward vector.

Best regards...

sublime1-ga

OK now we know what the net force on the ball is.  But the question
was what is the acceleration of the ball?  I gather from the equation
Fnet = m * a, that 'a' must be zero since you said Fnet is zero.  But
then the explanation seems a little circular.

Hi mathisfun-ga

You are correct and unfortunately your friend are wrong and
sublime1-ga has given you an absolut wrong answer.
This is an old problem argued in many physics books but you can see
the right answer here:
http://galileoandeinstein.physics.virginia.edu/lectures/gal_accn962.htm

"Throwing a Ball Upwards

To clarify ideas on the acceleration due to gravity, it is worth
thinking about throwing a ball vertically upwards. If we made another
movie, we would find that the motion going upwards is like a mirror
image of that on the way down-the distances traveled between frames on
the way up get shorter and shorter. In fact, the ball on its way up
loses speed at a steady rate, and the rate turns out to be ten meters
per second per second-the same as the rate of increase on the way
down. For example, if we throw the ball straight upwards at 20 meters
per second (about 40 mph) after one second it will have slowed to 10
meters per second, and after two seconds it will be at rest
momentarily before beginning to come down. After a total of four
seconds, it will be back where it started.

An obvious question so: how high did it go? The way to approach this
is to find its average speed on the way up and multiply it by the time
taken to get up. As before, it is helpful to sketch a graph of how the
speed is varying with time. The speed at the initial time is 20 meters
per second, at one second it's down to 10, then at two seconds it's
zero. It is clear from the graph that the average speed on the way up
is 10 meters per second, and since it takes two seconds to get up, the
total distance traveled must be 20 meters.
Speed and Velocity

Let us now try to extend our speed plot to keep a record of the entire
fall. The speed drops to zero when the ball reaches the top, then
begins to increase again. We could represent this by a V-shaped curve,
but it turns out to be more natural to introduce the idea of velocity.
Unfortunately velocity and speed mean the same thing in ordinary
usage, but in science velocity means more: it includes speed and
direction. In the case of a ball going straight up and down, we
include direction by saying that motion upwards has positive velocity,
motion downwards has negative velocity.

If we now plot the velocity of the ball at successive times, it is +20
initially, +10 after one second, 0 after two seconds, -10 after three
seconds, -20 after four seconds. If you plot this on a graph you will
see that it is all on the same straight line. Over each one-second
interval, the velocity decreases by ten meters per second throughout
the flight. In other words, the acceleration due to gravity is -10
meters per second per second, or you could say it is 10 meters per
second per second downwards.
What's the Acceleration at the Topmost Point?

Most people on being asked that for the first time say zero. That's
wrong. But to see why takes some clear thinking about just what is
meant by velocity and acceleration. Recall Zeno claimed motion was
impossible because at each instant of time an object has to be in a
particular position, and since an interval of time is made up of
instants, it could never move. The catch is that a second of time
cannot be built up of instants. It can, however, be built up of
intervals of time each as short as you wish. Average velocity over an
interval of time is defined by dividing the distance moved in that
interval by the time taken---the length of the interval. We define
velocity at an instant of time, such as the velocity of the ball when
the time is one second, by taking a small time interval which includes
the time one second, finding the average velocity over that time
interval, then repeating the process with smaller and smaller time
intervals to home in on the answer.

Now, to find acceleration at an instant of time we have to go through
the same process. Remember, acceleration is rate of change of
velocity. So, to find the acceleration at an instant we have to take
some short but non-zero time interval that includes the point in
question and find how much the velocity changes during that time
interval. Then we divide that velocity change by the time it took to
find the acceleration, in, say, meters per second per second.

The point is that at the topmost point of the throw, the ball does
come to rest for an instant. Before and after that instant, there is a
brief period where the velocity is so small it looks as if the ball is
at rest. Also, our eyes tend to lock on the ball, so there is an
illusion that the ball has zero velocity for a short but non-zero
period of time. But this isn't the case. The ball's velocity is always
changing. To find its acceleration at the topmost point, we have to
find how its velocity changes in a short time interval which includes
that point. If we took, for example, a period of one-thousandth of a
second, we would find the velocity to have changed by one centimeter
per second. So the ball would fall one two-thousandth of a centimeter
during that first thousandth of a second from rest-not too easy to
see! The bottom line, though, is that the acceleration of the ball is
10 meters per second per second downwards throughout the flight.

If you still find yourself thinking it's got no acceleration at the
top, maybe you're confusing velocity with acceleration. All these
words are used rather loosely in everyday life, but we are forced to
give them precise meanings to discuss motion unambiguously. In fact,
lack of clarity of definitions like this delayed understanding of
these things for centuries."



Where is the error of the answer? When you throw the ball there is two
forces so long the ball is in contact with your hand: the force F1
applied by your hand to the ball and the force of gravity, F2=mg. Both
applied to the ball. F1>F2 and the ball goes up. After the ball loss
contact with your hand exists only F2. This force is constant (for
hights small compared with the earth´s radius) and so is also the
acceleration. The inicial velocity  Vi pointed up (positive) decreases
because velocity and acceleration have opposing directions. After the
topmost point velocity and acceleration have the same direction and
the velocity increases until -Vi.
here are some plots:

Force (up positive)
|
|
|
|____________________________
|			     t
|
|__________________  (-mg)
|


acceleration (up positive)
|
|
|
|____________________________
|			    t
|
|__________________  (-g)
|
	

|velocity (up positive)
|
|\  (+Vi)
| \
|  \
|   \   (slope= -g)
|____\________________________
|     \			       t
|      \
|       \
|        \
|         \ (-Vi)



See also:
http://www.wsu.edu/~jtd/Physics205/Chap2.htm

I hope this help.  Forgive my poor english.

Francf

You are correct. Once the ball leaves your hand it enters free-fall.
Its motion, in classical terms, is a composition of its inertial
motion upwards and its acceleration downwards due to gravity. At all
points it is accelerating at 9.8 m/s^2. It is this constant
acceleration which continuously eats away the upwards speed,
eventually bringing it to an instantaneous halt and subsequently
increasing its overall speed downwards.

qed100-ga and francf-ga have it right.  Velocity is zero (for an
instant) at the top of the flight but acceleration and net force are a
downward constant value all the time between leaving your hand and
striking the ground.

I really hope you didn't pay anything for the original
answer; and the four star rating DEFINITELY needs to be
rescinded.  In both the answer and the "clarification"
there is a confusion between force and acceleration, for
example, the satement that the "force of gravity" being
9.8 m/s^2.  In the clarification, the misunderstanding
persists, since the "fine point" about discussion the
force rather than the acceleration is totally bogus; if
the net force is zero, then the acceleration is zero.  At
no point during the problem is the net force zero - not
when it is thrown, not when it's in the air, not at the top.

The second paragraph of the original answer is also in error.
The apparent weightless period in the airplane occurs during
the downward, free-fall portion of the plane's flight.  During
this time, the plane and the occupants inside accelerate at the 
same rate (9.8 m/s^2), providing an appearance of weightlessness
within the plane.  (Note that the plane's engines are still driving
it downward to cancel out the effects of air resistance.)

"The second paragraph of the original answer is also in error.
The apparent weightless period in the airplane occurs during
the downward, free-fall portion of the plane's flight.  During
this time, the plane and the occupants inside accelerate at the 
same rate (9.8 m/s^2), providing an appearance of weightlessness
within the plane.  (Note that the plane's engines are still driving
it downward to cancel out the effects of air resistance.)"

   Your comments about the official answer are largely correct, but I
think it's important to clarify something about your comment which
I've quoted above. It's true that on the downwards half of a parabolic
arc the passengers of an aircraft will be weightless, in fact every
bit as weightless as an astronaut on high orbit.

   But it's also true that, in principle, the same condition of
weightlessness, free-fall, can be had on the upwards portion of the
arc. If, hypothetically, we put the passengers inside a rocket on the
Moon's surface and launch it, upon engine cutoff the craft and
everything in it will be in free-fall, even as it coasts upwards. As
long as a projectile remains on a free-fall path as a function of
time, it'll be weightless.

   It's the same for an aircraft near Earth. As soon as it enters a
time-functional free-fall path, even though the airframe itself is
under stress due to both thrust and drag, and therefore is not
weightless, if the propulsion is adjusted to continuously keep its
mass center on that time-functional free-fall path, everyone inside
will be in free-fall, even on the upward portion.