Time taken to reach max/min height

[wadevala]

y for all real vertical throw, the time taken to reach max height must b shorter than time taken to return to starting point?

the answer given to me by a teacher is that in the upward motion, the max spd and deceleration are more so the time to travel the dist is shorter, and in the downward motion the max spd and acceleration are lower so the time taken to travel this dist is longer.

however i dun quite understand this explanation, can someone explain it more clearly to me..thx

 

[tiny-tim]

Hi wadevala!

If we ignore air resistance, your teacher is talking rubbish.

The path is is determined by a quadratic equation … it would be a parabola if it wasn't completely vertical … the speed at each point on the way up is the same as on the way down … if you took a film of it, and forgot which way round the film was, nobody could tell whether you were running the film forwards or backwards.

With air resistance, the ball is losing energy, so basically it'll go slower than it should on the way down, which means that the time to go down will be longer, but not for the reason he/she gave.

 

[ZapperZ]

y for all real vertical throw, the time taken to reach max height must b shorter than time taken to return to starting point?

the answer given to me by a teacher is that in the upward motion, the max spd and deceleration are more so the time to travel the dist is shorter, and in the downward motion the max spd and acceleration are lower so the time taken to travel this dist is longer.

however i dun quite understand this explanation, can someone explain it more clearly to me..thx

tiny tim is correct, and the fact that you are having a problem in understanding your teacher's explanation tells me that you are thinking of this correctly. So well done!

Zz.

 

[manjuvenamma]

Why is air resistance different in the two paths up and down? Shouldn't they be same since the velocities are just the same in both the paths?

 

[tiny-tim]

Why is air resistance different in the two paths up and down? Shouldn't they be same since the velocities are just the same in both the paths?

Hi manjuvenamma!

(you mean "the velocities are equal and opposite" or "the speeds are the same")

You're right that air resistance is the same up as it is down, for the same speed.

But the speeds aren't the same (at the same heights), because the (downward) acceleration is greater on the way up.

If air resistance is mAdx/dt, the the downward acceleration is g + Adx/dt; on the way up, that's more than g, and on the way down it's less than g; the greater the speed, the greater the difference.

 

[LURCH]

Why is air resistance different in the two paths up and down? Shouldn't they be same since the velocities are just the same in both the paths?

Yes, the air resistance is the same on the way up as it is on the way down. However, on the way up, the ball has just been thrown. On the way down, the ball has been in flight for some time. As Tiny Tim said, this resistance will change at different speeds, but it will continue to be present while the ball is in flight. So, if you imagine throwing the ball in a straight and level line, you know that it will continually slowdown due to air resistance throughout its entire flight.

The same is true for a vertical throw. Throw the ball straight up and down in a vacuum, and it will take the same amount of time to reach its maximum height as it does to return to the point from which it was thrown. But in an atmosphere, wind resistance will continue slowing the ball's progress all the way through its flight.

 

[manjuvenamma]

Hi Tiny Tim,

"But the speeds aren't the same (at the same heights), because the (downward) acceleration is greater on the way up."

The speeds should be the same at the same heights, and acceleration is also constant throughout the flight, that is what we assume and use in all our calculations. Of course, this is not considering air resistance. Now if you include air resistance, that will slow down the movement, but why do you think it will affect differently in the two paths up and down?

When so many of you are saying it is the way, I agree that it is most probably is the way, but I am not able to understand the cause or explanation you are giving.

 

[tiny-tim]

Hi manjuvenamma!

The speeds should be the same at the same heights, and acceleration is also constant throughout the flight, that is what we assume and use in all our calculations. Of course, this is not considering air resistance.

That's right!

Now if you include air resistance, that will slow down the movement, but why do you think it will affect differently in the two paths up and down?

Because it changes the acceleration:

If air resistance is mAdx/dt, the the downward acceleration is g + Adx/dt; on the way up, that's more than g, and on the way down it's less than g; the greater the speed, the greater the difference.

Have you done any problems in which there are two weights M and m connected by a rope which goes round a pulley?

Each weight moves with an acceleration of g times ±(M - m)/(M + m) … until the rope breaks, or one weight hits the ground!

Similarly, air resistance means that the acceleration is no longer g … and it's different on the way down from the way up!

 

[manjuvenamma]

Hi Tiny Tim,

I am a beginner in Physics. I found joy in solving kinematics problems and predicting horizontal ranges, max heights etc etc. where throughout I assumed g is constant. Now also I think we should say g is constant but now we have to take into account another factor i.e air resistance which we all know is real and exists.

How that affects the velocity of the stone is the questions. I believe g is independent of the air resistance and does not get affected.

Now, I know nothing about air resistance beyond my guess that that is some kind of friction and slows down the relative motion in both the ways. You seem to be saying some thing different with an equation.

I have not yet done anything with pulleys.

I will try to understand explanation after I finish with your pulley analogy. I win my bread with software development, so it takes a bit of time to absorb your explanation. Thanks for your time and patience.

 

[tiny-tim]

Hi manjuvenamma!

Now also I think we should say g is constant but now we have to take into account another factor i.e air resistance which we all know is real and exists.

How that affects the velocity of the stone is the questions. I believe g is independent of the air resistance and does not get affected..

Yes that's right, of course … g itself is not affected.

And g is independent of speed, while air resistance obviously isn't.

So when you're making calculations (as opposed to just a rough description, as here), you do have to treat them separately!

Now, I know nothing about air resistance beyond my guess that that is some kind of friction and slows down the relative motion in both the ways. You seem to be saying some thing different with an equation.

I have not yet done anything with pulleys.

Actually, my equation is the usual friction equation … the force of friction is a constant (usually written µ, pronounced "mu") times the velocity (and directly opposite to the velocity). So the extra acceleration from friction is -µdx/dt.

Air resistance is slightly different at low speeds, basically because the air gets dragged along with the object, and very different at higher speeds (think of terminal velocity, and turbulence round wings).

For pulleys, I suggest you do a forum search for "pulley", and try some examples!

 

[indr0008]

let's just go like this: gravity: 10 m/s^2, friction: (assume constant) 2 m/s^2

so when the ball was moving upwards, the acceleration was: 12 m/s^2 slowing it down
when the ball was moving downwards, the acceleration was 8 m/s^2 speeding it up downwards

and because the distances traveled on both ways are the same, call it H, from the simple equation H=(a*t^2)/2 we can see that the time taken would be longer for the downfall, right?

i knew that this approximation is the silliest thing ever mentioned in this thread, but i hope it helps to visualize the problem