Starship Asterisk*

by **BDanielMayfield** » Tue Jul 24, 2018 2:58 am

I didn't notice this quiz until after the voting had closed, but in all honesty I too would have gone with the overwhelming majority who selected "same". But, on the bright side, this result reaffirms my skeptism about just accepting "consensus" opinions.

Bruce

by **Guest** » Sun Jul 22, 2018 7:34 pm

If a ball is bounced hard enough, can it travel faster upward than it does if dropped downward from the same height? In other words, if I dribble a ball, after I bounce it will it come to my hand faster than it dropped from my hand( let’s just say it’s the same distance )

by **wonderboy** » Fri May 21, 2010 3:30 pm

To me this is simple.

Of course the ball is going to go faster upwards than it is downwards. This is because when you throw the ball upwards you are using a force greater than that of gravity (initially) to make the ball rise. on the way down, there is no initial force. Both ways the ball will encounter drag which will slow it down, so therefore a ball rising with a force greater than that gravity will rise quicker than a ball which falls relying solely on gravity.

Paul

by **neufer** » Thu May 20, 2010 8:07 pm

Hold a standard plastic soda straw crosswise on the middle of your four fingers with your thumb.

Then turn your palm over and toss the straw forward overhand as fast as you can
so as to give it a strong backspin as the straw rolls forward off your fingertips.

Does the straw take longer to go up or to come down?

by **basic_user** » Thu May 20, 2010 5:55 pm

It depends. Consider a hand on a table with a ball sitting on the hand, and then the hand and ball moving up from the table. Assume the hand moves at a constant acceleration and the ball and hand are inelastic. Say the hand stops a foot above the table and the ball continues to rise. From this point above the table, ignoring air resistance, the ball takes as long to get to the top of its flight as to return to this spot, and its upward and downward velocities at this point are the same. The variable is what happens in the first foot. The falling ball crosses the distance at 1 g and comes to an immediate stop. The time for the upward rising ball to cross one foot depends on the acceleration of the hand. Generally, I would expect the hand to accelerate much faster than 1 g, so up would be faster than down. (If this weren't true we wouldn't be able to catch a ball that falls out of our hand.)

by **mndt** » Thu May 20, 2010 4:28 pm

On the start of the way up the ball has a certain amount of energy.
k=(1/2)mv^2

But friction with air makes it lose some of it's energy. so K (on the way down) < K (On the way up)

Since m (mass) is the same. V (Velocity) should be less. so it takes more time for it to come down.

by **neufer** » Wed May 19, 2010 6:55 pm

Non self propelled objects that generally come down faster than they went up:

1) Lawn Darts

2) Paper airplanes

3) Balsam/Styrofoam gliders set for a slow stall

4) Kites

by **neufer** » Wed May 19, 2010 2:58 pm

RJN wrote:The poll is now closed and the correct answer is now included in the initial post at the top. - RJN

Do these results differ significantly from the pre-open discussion poll results from last week?

I can understand people voting for "the same"
but I am rather baffled by the number of votes for "up".

(All folks had to do was to wad up a piece of paper and toss it.)

by **RJN** » Wed May 19, 2010 2:41 pm

The poll is now closed and the correct answer is now included in the initial post at the top. - RJN

by wh » Wed May 19, 2010 2:08 am

To me, the force of air resistance is in both directions against the motion of the ball, and is in both directions proportional to the velocity at any point. Whatever the initial velocity, that is reduced by constant gravity, which is the same in both directions. However high the ball goes, it is from that height that it will have time to go down with the same acceleration as deceleration, with the same proportional drag against that motion. Ergo, aequum.

.

by **dog** » Tue May 18, 2010 10:34 pm

When the ball moves in the atmosphere, some of the air is pushed along with it (drag). As the ball continually slows down on its upward flight, some of this moving air will push on the ball. When the ball reverses direction and falls, it must now move through this rising stream of air. The ball must take longer falling as it must also overcome the greater air distance and resistance.

by **neufer** » Tue May 18, 2010 9:03 pm

Bert and Ernie wrote:If going up reaches escape velocity, it takes longer to go up.

If going up reaches escape velocity, please call the Washington Nationals immediately!

by **Bert and Ernie** » Tue May 18, 2010 8:42 pm

If going up reaches escape velocity, it takes longer to go up.

by **neufer** » Tue May 18, 2010 7:42 pm

[list]ƒ = integration

T_u = total time up
T_d = total time down

V_u(z) = Velocity up at a given height z
V_d(z) = Velocity back down at a given height z
V_u(z) ≥ V_d(z) due to any air resistance in the interim

T = ƒ dt = ƒ dz/(dz/dt) = ƒ dz/V(z)

T_u = ƒ dz/V_u(z) : T_d = ƒ dz/V_d(z)

V_u(z) ≥ V_d(z) due to any air resistance in the interim

Thus: 1/V_u(z) ≤ 1/V_d(z)

Ergo: T_u ≤ T_d : { T_u = T_d if and only if air resistance = 0 }[/list]

by **Amir** » Tue May 18, 2010 6:40 pm

at first i voted for "Same", cuz i assumed that the ball is thrown in vacuum. but when you all said that it's in atmosphere, i took another look: "...in the air..."! so i'll say "down".
i agree with Chris & biddie67:

Chris Peterson wrote:...Viewed qualitatively, drag is acting downwards while the ball is rising, which reduces the time the ball spends traveling upwards; drag is acting upwards while the ball is falling, which increases the time the ball spends traveling downwards. ...

we throw the ball so we give it a Kinetic energy (K₁), which drag force (fk₁) will reduce it while it changes to Potential energy (U): U = K₁- fk₁
when the ball is coming back drag force will do the same thing again while Potential energy changes to Kinetic energy (K₂): K₂ = U - fk₂
thus K₂ < K₁, same mass --> V₂ < V₁, same distance --> t₂> t₁

by **WrB** » Tue May 18, 2010 6:29 pm

WrB wrote:Think of the same situation in a vacuum. The balls initial velocity determines the balls maximum height (from the equation of movement in a vacuum). I would presume that it would be equal time in this case.

But in atmosphere, the ball has drag and the ball never reaches this calculated height (from our equation of movement in a vacuum). Some of that kinetic energy is lost due to drag. At the top of the arc, the balls potential energy is not equal to the balls initial kinetic energy because of these losses. On the way down, as the ball crashes into the ground, the balls downward speed will be less than the balls initial upward speed, due to the drag exerted on the ball during both the upward and downward legs of this trip. Thus the average speed upward is greater than the average speed downward. So I would say that it takes longer for the ball to fall.

Correction: "as the ball crashes into the ground" needs to be "as the ball approaches it's initial height", due to the fact that we are not throwing the object from ground level, but slightly above it.

by **WrB** » Tue May 18, 2010 6:12 pm

Think of the same situation in a vacuum. The balls initial velocity determines the balls maximum height (from the equation of movement in a vacuum). I would presume that it would be equal time in this case.

But in atmosphere, the ball has drag and the ball never reaches this calculated height (from our equation of movement in a vacuum). Some of that kinetic energy is lost due to drag. At the top of the arc, the balls potential energy is not equal to the balls initial kinetic energy because of these losses. On the way down, as the ball crashes into the ground, the balls downward speed will be less than the balls initial upward speed, due to the drag exerted on the ball during both the upward and downward legs of this trip. Thus the average speed upward is greater than the average speed downward. So I would say that it takes longer for the ball to fall.

by **MurryClan** » Tue May 18, 2010 5:55 pm

It is not said how the ball is thrown. If it is thrown by me, The ball would not have enough speed to reach an altitude to obtain terminal velocity. So I figure it would take the same time going up as down. As I said it was not stated how the ball was thrown and one could make a way to have that ball thrown at speeds greater than terminal velocity, in which case it would take the ball longer to come down.

by **biddie67** » Tue May 18, 2010 5:45 pm

I think that it will take longer to come down. At each point in altitude, the combined forces of gravity and drag will be the same on both the trip up and the trip down (unless the barametric pressures change inbetween the two trips). It seems that the big factor is the force used to toss the ball up to start with. It will go up at the velocity (and length of time embedded in it) as it overcomes gravity and drag and is exhausted of that residual of energy. But on the way back down, it would seem to be only at the mercy of the forces of gravitational acceleration which might be less than the original force to toss it so the time to come back down would be longer ....

by **Hoebeau** » Tue May 18, 2010 5:40 pm

Without any further information, you must consider the scenario where the ball is thrown upwards faster than terminal velocity. When it stops rising under gravity's downward pull, it will then accelerate downward until it reaches terminal velocity, which is slower than its starting upward velocity in this scenario. Thus, it will take longer to fall the same distance it rose.

by **BMAONE23** » Tue May 18, 2010 5:15 pm

Perhaps it is time to get the mythbusters involved Or Nolan Ryan and a Radar speed detector.

by **Chris Peterson** » Tue May 18, 2010 5:02 pm

Feagles wrote:Clearly, the answer depends on how the ball was thrown.

I don't think it matters at all how the ball is thrown. Regardless of how fast or how slow the ball is launched, it will always take longer to come down than it did to go up.

by **Feagles** » Tue May 18, 2010 4:37 pm

Clearly, the answer depends on how the ball was thrown. If it was catapulted faster than it's terminal velocity on return, it will take longer to come down. If less than terminal velocity it should be the same. I was tempted to add that where it was being thrown could enter in as well. If it's on the space shuttle .... But since up and down are so subjective/irrelevant I decided to let that go. On the other hand has anyone checked to see if one could hand toss a ball on a lower gravity celestial body with an atmosphere fast enough to reach terminal velocity there? But that atmosphere would probably not be called air.

by **Beta** » Tue May 18, 2010 4:07 pm

BMAONE23 wrote:The higher [the ball] travels, the more likely it will reach terminal velocity on its return trip. Since no human can throw a ball at 200+ mph it would likely return faster if the thrown altitude allows for a faster return speed.

You are imposing a limit on how fast one can throw a ball, but no limit on how high one can throw it.

The ball cannot return with more energy than it had when it started. If you can't throw it that fast, you can't throw it high enough to come back that fast.

by **Chris Peterson** » Tue May 18, 2010 3:56 pm

waywardson wrote:i thought of it as an energy problem. as the ball travels on its arc, air resistance is constantly being reduced, which means that it's maximum kinetic energy is reduced both the start and end of the trip down being lower than the trip up...

The problem becomes more complex if you add a horizontal velocity component. I assumed that the ball is thrown straight up, so everything happens along a single axis (i.e. no "arc", unless it is an arc of position versus time).

Starship Asterisk*

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