Do you know Gravity ?
If you throw a small ball vertically upward in real air with drag, does it take longer to go up or come down?
Labels: funphysics
posted by Rajesh Lal at
9:18 AM
Thursday, August 6, 2009
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18 Comments:
the small ball takes longer to go down then to go up, becaus eof terminal velocity, (if the ball reaches it) the ball can only go so fast. but if u throw it then u have the speed of the throw
this prolly isnt right, but i just felt like trying. :D,
I think it takes the same amount of time both ways.
Terminal Velocity is not a constant. It dependant on the amount of drag an object may have. less drag can achieve higher terminal velocity. But that terminal velocity would also hinder the upward speed, if your strong enough to even achieve it, which your not.. lol
So the drag should be the same both ways, and gravity should slow it down at the same rate that it accelerates it on the way down.
That is correct. It is the same both ways. Physics 101.
it will take the same amount of time up and down unless terminal velocity is exceeded on the way up, while this may be dificult to do with a baseball, it would be easier with a beach ball.
It could be argued that it would take longer to come back down because the question doesn't say you catch the ball, so the ball has a further distance to travel on the way down than it did on the way up.
Or... if you were able to throw the ball high enough to escape gravity, which the question does not directly forbid being able to do, then it would take an infinite amount of time longer to come back down to OUR planet than it did to go up!
I tested this, and noted it ALWAYS finishes going up before it finishes coming down no matter how high I throw it.
I thought initially it should be the same (considerations about terminal velocity aside), but what about air turbulence?
On the downward portion of the throw the ball would be falling through air that had been disturbed on its upward path. My intuition is that the air pressure would be slightly lower because it would have been pushed out of the way on the way up, and therefore the ball would fall slightly faster. The larger the ball and the faster it is thrown the greater this effect would be.
Air distubances can cause surprisingly large effects. In New York a few years back a plane crashed by flying through wing tip vortices caused by a previous plane. They are like little whirlpools of air and go on for miles.
The gravitaional force is always puling the ball down
and is constant (if assume the height is small compared
with the Earth's radius).
The drag always reduces the speed.
So when the ball is going up it experiences a force
mg+D and when it is coming down, it experiences a
smaller force mg-D. m, g and D are positive quantities.
Ignoring the complications of precisely what D is, the
ball will be decelerated more strongly going up than it
is accelerated when coming down.
As the ball has to travel the same distance in both
directons, it must take longer to come down than to go up.
SP33D - your intuition is correct.
Mister Fahrenheit - quite a clever idea there, but at
best you could only hope to reduce D to 0 on the way down.
Ragknot - I like your humour.
lol, thank you Chris, haha that made my day. lol
Both ways up and down are equal to each other, but at the same high above earth level back way velocity is less of the first way cause of drag. Therefore back time is longer.
If you or I would throw a ball up, The ball would be traveling faster than 9.8 m/s^2 and gradually climb slower and slower until the max, at 0 m/s^2 and then fall. The ball can only fall as fast as gravity pushes it. So, as long as you throw the ball up faster than 9.8m/s^2, then the ball going up would be faster than the ball comming down.
Last ANonymous, too be blunt, you are talking gibberish. How can you throw a ball with a speed measured in units of acceleration?
Also, you nowhere mention drag.
I'll add that I'm not overly proud of my previous answer. I should have mentioned the equation v^2 = u^2 + 2as. I didn't as it assumes a (and hence D) would be constant, and that is definitely not true.
For very low speeds D is proportional to v, and for low speeds, D is proportional to v^2. After that it becomes complicated.
I'm a nit. Of course I can use v^2 = u^2 + 2gh
g is accel due to gravity, d is accel (retardive) due to drag
and h is the maximum height the ball reaches.
For simplicity, I assume the same value of drag throughout.
I'm keeping to +ve values as far as possible.
Ball going up: v^2 = 2(g+d)h => h = (v^2)/(g+d)
Ball coming down: u^2 = 2(g-d)h => h = (u^2)/(g-d)
Equating and changing to ratios => (v^2)/(u^2) = (g+d)/(g-d).
Time taken is inversely proportional to speed, so
(Tdown/Tup)^2 = (g+d)/(g-d)
As g and d are +ve, (g+d) > (g-d), so Tdown > Tup
I think we are getting much too deep here. Once the ball leaves the hand it is being acted on only be gravity and air friction. Air friction does not care about the direction, so the drag would be greatest at the bottom and least at top, just as the speed would. It should be the same up as down.
Okay, want to change my answer. If drag is always against the movement, it would be slowest coming down. It would slow it up going up and continue to slow it up coming down(follow the ups and downs?).
It would be slower going up. The drag + gravity + force needed for it to go up is lower than the gravity pulling it down.
size of the ball has nothing to do with it, it's the mass that matters
The size of the ball affects the drag, it is a crucial part of the solution to the problem.
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