[Guest]

LIS's little cannonball problem reminded me of this one.

You throw a ball up in the air.

It goes up. It comes down.

You catch it at the same height you threw it from.

Now, did it take longer going up, or coming down?

Don't forget air resistance

[Guest]

going up?

[Brandonb]

Assuming that the wind up for the throw is not counted as part of the airtime, then the ball starts from zero to instant momentum upward. It battles gravity and air resistance as it goes up, then slows down to zero momentum. At that point it begins decent from zero and gravity causes it to accelerate at the same rate that it decelerated on the way up. It battles the same wind resistance on the way down and hit zero acceleration exactly as when it was released. Because the wind resistance, if remaining a constant, and the spin on the ball is constant, and gravity of course is constant. Then I'm pretty sure it's the same both ways.

[Guest]

hahaha, I actually tried throwing my remote into the air for this one haha

if you threw it up with greater force than the gravitational force then wouldnt it be quicker upwards as it would have more speed?

[Guest]

hahaha, I actually tried throwing my remote into the air for this one haha

if you threw it up with greater force than the gravitational force then wouldnt it be quicker upwards as it would have more speed?

If you throw it up with greater force than gravity it takes longer going up cos it goes into orbit . Just kidding, I might not answer this one, and let everybody sort it out between themselves.

[Guest]

i guess i didnt mean force i meant so the initial velocity was faster than the terminal velocity made by gravity lol, got nice images of balls orbiting in space now haha

[Guest]

It takes longer coming back down.

While going up, gravity and air resistance work in the same direction (downward) on the ball, providing deceleration (or acceleration in the downward direction). On the way down, gravity still pulls downward, the air resistance is now pulling upward. This results in a lower net force on the ball, and therefore a lower acceleration in the downward direction. A lower acceleration means that the ball takes a longer time to travel the same distance that it did going up. So it takes longer to come back down.

We can also think about the ball's starting velocity. If the ball's starting velocity is greater than the ball's terminal velocity (the maximum velocity an object can reach in free fall), then the ball would definitely take less time on the way up than on the way down, since it's speed on the way down is capped.

Edited May 26, 2008 by flowstoneknight

[Guest]

BETTER NOT FORGET THE GRAVITY

THE TIME TAKEN ARE SAME

[Guest]

It takes longer going up I think because as it goes up, right before it reaches the peak, it has to slow down, whereas when it's coming down it gains speed the whole time.

[Guest]

It's the same. It starts going up fast, slows to zero. Must accelerate from zero back down. It'll land in your hand with the same final velocity (in the opposite direction) as when you released it. Air resistance, if it has an uneven effect (which I'm not convinced it does), is tiny and can be ignored. Think of the graph of the line S=.5at^2 + vt, where S = y (the height at time t), a = 9.8 m/s/s, v = however hard you threw it, and t = x. The peak of the resulting parabola (y = 4.9 x^2 + vx) is directly between the two x-values where y=the height of your hand. Thus the time spent getting to and coming down from that peak is the same.

[Guest]

It takes longer going up I think because as it goes up, right before it reaches the peak, it has to slow down, whereas when it's coming down it gains speed the whole time.

The ball is slowing down the whole time while going up, since (I assume) we're counting the "height that you threw it from" as being the height at which you release the ball.

[Guest]

More simple than cannons - or is it?

Just to help I think the rate of free fall is 9.81m/s/s at sea level, but takes a while to pick up speed due to air resistance which (I think) is the same as goin up! I would be inclined to ignore air resistance it is probably relevant to different objects but not the same size shape and weight. In this case it is the self same object and octopuppy loves red herrings, I think they are a bit salty!

Q how fast can you through the ball in the air?????

Does it matter???

[itachi-san]

More simple than cannons - or is it?

Just to help I think the rate of free fall is 9.81m/s/s at sea level, but takes a while to pick up speed due to air resistance which (I think) is the same as goin up! I would be inclined to ignore air resistance it is probably relevant to different objects but not the same size shape and weight. In this case it is the self same object and octopuppy loves red herrings, I think they are a bit salty!

Q how fast can you through the ball in the air?????

Does it matter???

also Q how far can you throw it vertically? and does that matter?

[Guest]

also Q how far can you throw it vertically? and does that matter?

How far meaning how high (should be the same) are you meaning allowed to throw it or theoretically as far as supeman can throw?

try two

1 as high as you can (50m)

2 just 9.81 metres or even 2 metres

[Guest]

It has to take the same time. Simple physics. The air resistance is negligable, at each point in both the rise upwards and down the air density will be equal at each certain height, negating it. Same thing happens with the acceleration: it's -9.8m/s/s up and 9.8 m/s/s down, meaning that so long as distance remains equal (and it does) it will be the same. The physics are quite simple actually, drag does not need to be calculated.

[Guest]

i agree with the physic that say acceleration is the same and air resistance is the same, however coming down there is a terminal velocity of whatever it is for the ball, but going up depends how fast your throw is?

[Guest]

It has to take the same time. Simple physics. The air resistance is negligable, at each point in both the rise upwards and down the air density will be equal at each certain height, negating it. Same thing happens with the acceleration: it's -9.8m/s/s up and 9.8 m/s/s down, meaning that so long as distance remains equal (and it does) it will be the same. The physics are quite simple actually, drag does not need to be calculated.

I agree. Although if you wanted to bring air resitance into it then you could say that air fesistance on the way up would have to be the same as air resistance on the way down (assuming the object is a standard symetrical shape, a sphere is best I think). This would increase the acceleration on the way up, and decrease the acceleration on the way down by the same amount. Effictively cancelling each other out. But air resistance (as stated before) is so neglible it would only make any noticeable difference if we could throw incredubly high, but that won't be happening.

The only thing I can think of that would cause going up to be slower than coming down would be the shape of the thrown object. If a yramid was thrown point upwards it would go up faster than it would come down.

[Guest]

To people saying that it takes the same amount of time, here's an explanation:

Let's look at the forces involved. We can't ignore air resistance since it's specifically mentioned in the OP. So the two forces acting on the ball's "flight" are air resistance and gravity. It is important to know the directions in which these forces are acting during the two parts of the ball's journey (up and down). We know that gravity pulls downward in both parts. Air resistance works in the direction opposite that of the ball's motion (hence resistance), so it pulls downward while the ball is moving up and upward while the ball is moving down. Here's a simple diagram of the two forces involved:

The force of gravity (we'll assume it's constant and that the ball isn't thrown to astronomical heights) is the same whether the ball moves upward or downward. But since air resistance works in different directions during the two parts, the net force on the ball while going up and going down are not the same. The ball experiences a greater force while going up (and therefore a greater acceleration in the downward direction) than it does while going down. This difference in acceleration results in a difference in the time taken to travel the same distance.

[Guest]

To people saying that it takes the same amount of time, here's an explanation:

Let's look at the forces involved. We can't ignore air resistance since it's specifically mentioned in the OP. So the two forces acting on the ball's "flight" are air resistance and gravity. It is important to know the directions in which these forces are acting during the two parts of the ball's journey (up and down). We know that gravity pulls downward in both parts. Air resistance works in the direction opposite that of the ball's motion (hence resistance), so it pulls downward while the ball is moving up and upward while the ball is moving down. Here's a simple diagram of the two forces involved:

The force of gravity (we'll assume it's constant and that the ball isn't thrown to astronomical heights) is the same whether the ball moves upward or downward. But since air resistance works in different directions during the two parts, the net force on the ball while going up and going down are not the same. The ball experiences a greater force while going up (and therefore a greater acceleration in the downward direction) than it does while going down. This difference in acceleration results in a difference in the time taken to travel the same distance.

Nice try, but you're forgetting that on the way up, you threw the ball. The only force acting on the ball on the way down is gravity, accelerating from zero. My answer is correct (edit: as long as it is a symmetrical object). We can talk about impulses and everything, and drag, and acceleration, but you will find that the speed at which the ball leaves your hand will be the same speed it hits your hand at.

Edited May 27, 2008 by Sky

[Guest]

Nice try, but you're forgetting that on the way up, you threw the ball. The only force acting on the ball on the way down is gravity, accelerating from zero. My answer is correct (edit: as long as it is a symmetrical object). We can talk about impulses and everything, and drag, and acceleration, but you will find that the speed at which the ball leaves your hand will be the same speed it hits your hand at.

Because air resistance doesn't work on balls that are thrown?

[Guest]

Air resistance is the same in both directions though. It cancels itself out. It affects both the ball going up and the ball going down. It's not like its stronger in one direction.

[Guest]

Friction is a reaction force. It only exists when another force is present. Just so you know.

[Guest]

Friction is a reaction force. It only exists when another force is present. Just so you know.

Yes, and that "another force" is the force exerted on the air molecules by the ball as it pushes them out of the way. This is why air resistance always works in the opposite direction of the object's motion. Please go and read my explanation again and if you still aren't convinced, please point to where you think I went wrong.

[Guest]

To people saying that it takes the same amount of time, here's an explanation:

Let's look at the forces involved. We can't ignore air resistance since it's specifically mentioned in the OP. So the two forces acting on the ball's "flight" are air resistance and gravity. It is important to know the directions in which these forces are acting during the two parts of the ball's journey (up and down). We know that gravity pulls downward in both parts. Air resistance works in the direction opposite that of the ball's motion (hence resistance), so it pulls downward while the ball is moving up and upward while the ball is moving down. Here's a simple diagram of the two forces involved:

The force of gravity (we'll assume it's constant and that the ball isn't thrown to astronomical heights) is the same whether the ball moves upward or downward. But since air resistance works in different directions during the two parts, the net force on the ball while going up and going down are not the same. The ball experiences a greater force while going up (and therefore a greater acceleration in the downward direction) than it does while going down. This difference in acceleration results in a difference in the time taken to travel the same distance.

Hmm... I see what you mean through the air resistance bit there, but at low heights, it should still be negligable. Maybe at a higher speed and height... air resistance would greatly affect it then...

[Guest]

Indeed, air resistance would provide no significant difference in most cases. I only insist on including it because it is specifically mentioned in the OP. A part of making any physical model is choosing what to ignore. It's just that in this case, I don't think air resistance is something to ignore. So I guess our disagreement is ultimately on what the question was asking.

As an aside, the title "what goes up must come down" is often used for physics problems involving this question, or similar variations.

[Guest]

also Q how far can you throw it vertically? and does that matter?

Although I'm not sure this was asked seriously, I'll answer anyway. It doesn't matter how hard you throw it or how high you throw it (both are essentially the same thing, because the harder you throw it, the higher it goes...).

I only insist on including it because it is specifically mentioned in the OP.

Just because something is mentioned in the OP doesn't mean it's important.

You throw a ball up in the air.

The real question is, why did you eat the ball in the first place? And why did you catch it after you puked it up?