Baseball Toss
If you are holding a baseball ... and you toss it straight up, will the upward flight or the downward flight take longer? ... and why?
posted by Zaux at
2:21 PM
Friday, February 5, 2010
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28 Comments:
the time it takes to go up should be the same as time it takes to fall back to the same position at which it left your hand
Distance travelled (x) under constant acceleration (a) in time (t) is calculated from x=1/2at².
Gravity is a constant force of acceleration and the equation makes no distinction between acceleration or deceleration (which is simply negative acceleration) so the time is only dependent on two things: distance and accelerating force.
These two variables are the same in the upward, and the downward path of the ball, so the time will be the same in both directions.
Because of air resistance, it would take longer coming down, than going up.
Chris...wouldn't there be air resistance on the way up as well?
On the way up mg is pointing down and drag is pointing down. On the way down, mg still points down. but drag points up. So there are larger forces on way up than on way down. More force, same mass, greater acceleration, so things happen quicker on way up.
I'll knock up an idealised calculation in a few minutes.
h = ½ a t². Choosing convention so that all quantities are > 0. Also taking d to be a constant magnitude (else the math gets messy).
h = ½(g+d)t_u² = ½(g-d)t_d²
=> (t_d/t_u)² = (g+d)/(g-d) > 1 (d > 0).
=> t_d > t_u (if have drag).
Chris,
On the way up the overriding force acting on the ball is the kinetic energy given to it by the thrower, this force (acting in the upward direction) is the one pushing against the drag acting in the downward direction.
In your scenario there is no upward force on the upward journey.
DualAspect. You cannot equate energy to force. The thrower provides kinetic energy to the ball, which is then lost as the ball gains height and overcomes drag. Once the thrower has let go, he is no longer providing any force. The only time there is an upward force is while the thrower is actively accelerating the ball in his hand.
Acoording to ....
It takes longer to drop than to fly up. The ball works against air resistance on the way up and continuously loses energy. Thus the total energy of the ball at a point on its way up is greater than its energy at the same height on the way down. Since the potential energy (energy due to its height) is the same at both instances, the difference in energy must be due to a reduced kinetic energy. Thus the falling ball is moving more slowly and will take more time to cover the same distance.
At least, that's what the solution states ... just wondering, if it loses energy on the way up, why is the total upward energy greater than the downward energy at the same point in flight?
Onthe way up drag assists gravity in decelerating the ball and so the effective g is g+d > g. On the way down, drag opposes gravity and so the effective g is g-d < g.
Hi Zaux. The ball is going down after it had been going up. So it has had longer to lose energy to drag.
ok ... thanks
Sticking with the approximation that the drag is constant in magnitude, to get to a height h, a KE = mgh + dh is required. But when the ball gets to h, it only has potential energy mgh, dh having being lost. On the way down, it gains KE = mgh via gravity, but loses energy dh via drag. On the way up the average KE and hence speed is higher than on the way down. Hence it takes longer to come down. The math I posted at 3:16 PM really covers everything you need to get the result.
I expect most of you know that you can punch an (air-filled) party balloon upwards quite fast, but it comes down quite slowly.
I concede!
My head still says that it's not true but it's getting late (UK), my physics 'A' Levels are too long ago to work it out and I can't argue with the logic.
now .. that's a good simplified example
It's a bit of a naughty argument, because in that extreme case, buoyancy effects are significant. But the bouyancy always opposes g. So let b be the buoyancy (force). In all of the equations and other blather, replace g with (g-b), regardless of direction of motion.
It did not say you caught the ball.
Therefore it will hit the ground, or floor, a longer path downward.
So the downward path takes longer.
DualAspect, don't give up. The equation I used: h = h0 + u t + ½ a t², but sneakily I did it backwards, so that u (the initial velocity) = 0. i.e. I solved the upward part as if I was watching it in a film running backwards. It just keeps the equations neater - it doesn't change the physics.
If you've done calculus, this should be easy to follow.
a = dv/dt => dv = a dt.
Integrating => v - u = a t => v = u + at
But v = ds/dt => ds = v dt = (u + at) dt
Integrating => s = s0 + ut + ½ a t², where s0 is the starting position. I cunningly chose initial conditions so that u = 0 and s0 = 0, and chaged s to h.
But the thrower was in a pit 20 feet below the ground, and the ball landed above him (on the ground) and so it was much quicker coming down ;)
I saw something similar on Myth Busters, with a little twist. A bullet was shot into the air with a gun straight up. Different methods of measurements were taken, but conclusive results were obtained. The bullet would reach terminal velocity falling. Falling always took more time.
They also looked into injuries caused by falling bullets, very little was found. A ball is not a bullet, the velocities can be much different, but I would think they would be similar.
Good point. There is essentially no limit to how fast the ball can be thrown upwards, but it will be limited by its terminal velocity on the way down.
The upward force is variable, u can throw it faster or slower compared to the downward speed. The downward motion completely depend on gravity and mass of ball, so it wont change with respect to the force with which it was thrown up. But the upward speed can change, so final answer is indeterminate.
It would take longer coming down do to the pendulum effect. However gravity makes it so it takes longer going up. Maybe. I could be wrong, but based on what I know this is my answer
From The Godson
This post has been removed by the author.
RiDiM and The Godson. The only consolation I have for you is that I have seen more appalling responses to questions. Read some of my posts (above) if you want to gain any understanding of why the ball takes longer to come down.
This post has been removed by the author.
Hi again RiDiM and The Godson. Sorry about the bluntness in the last post. I was been tongue-in-cheek for the initial blasting I gave.
If it's any consolation, most of the posters here, are rubbish at physics too. (Sorry folks, but it's true). Although I simplified my equations by letting the drag/wind-resistance be constant, the drag is roughly proportional to the speed (at low speeds). {I have posted a fuller explanation in the "Raindrops keep fall..." problem).
Visualise punching a party-balloon into the air, it goes up very (surprisingly) fast and then comes down at almost constant speed.
I don't know what a pendulum has got to do with the problem, unless you're imagining a feather fluttering down a bit like a pendulum bob would swing. Forget about that, that's difficult to analyse.
Gravity contributes in the same way going up as going down. If it weren't for drag, it would take the same time both ways. The drag reduces the speed during the entire proceedings. So the ball's speed is reducing all the time. So it is going slower on the way down than on the way up. So it takes longer coming down.
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