Sandy Koufax throws a fast ball... perfectly straight and level....
January 22, 2016 8:26 PM Subscribe
At the same instant, The three year old fan drops the autographed ball Sandy gave him before the game from the same exact height that Sandy released his pitch. Which one will hit the ground first?
I made up this hypothetical reality because I am confused by what I read in a science workbook. I am no scientist but I find myself fascinated by science and the study of our physical reality. I am reaching out for physics gurus to make the following understandable to me; "An object that is thrown is called a projectile. A projectile that is thrown horizontally will land on the ground at the same time as an object that is dropped."
I understand that gravity acts on both balls equally pulling them with the same force toward the ground. They are of equal weight and mass. However Sandy's ball has been impelled from it's inertia by the force of Sandy's throwing motion. This means that x amount of newtons move it forward towards where it will strike the ground when gravity pulls it there. Will this not cause it to take a longer amount of time to hit the ground than the autographed ball that simply falls directly from being released from the restraining grasp of the child? It just seems to my mind that due to the forward force the pitched ball will stay in the air longer than the ball falling straight down. If I am missing something here help me understand what it is?
I made up this hypothetical reality because I am confused by what I read in a science workbook. I am no scientist but I find myself fascinated by science and the study of our physical reality. I am reaching out for physics gurus to make the following understandable to me; "An object that is thrown is called a projectile. A projectile that is thrown horizontally will land on the ground at the same time as an object that is dropped."
I understand that gravity acts on both balls equally pulling them with the same force toward the ground. They are of equal weight and mass. However Sandy's ball has been impelled from it's inertia by the force of Sandy's throwing motion. This means that x amount of newtons move it forward towards where it will strike the ground when gravity pulls it there. Will this not cause it to take a longer amount of time to hit the ground than the autographed ball that simply falls directly from being released from the restraining grasp of the child? It just seems to my mind that due to the forward force the pitched ball will stay in the air longer than the ball falling straight down. If I am missing something here help me understand what it is?
However Sandy's ball has been impelled from it's inertia by the force of Sandy's throwing motion. This means that x amount of newtons move it forward towards where it will strike the ground when gravity pulls it there.
It would be tidier to say, Sandy's ball has been accelerated horizontally and now has a non-zero horizontal velocity 'V_hor_initial = 30 m/s', while both balls still have zero vertical velocity.
Upon release, gravity accelerates both balls downward. After a time t, the vertical component of both balls' velocity is V_vert = g*t.
So at t = 1 second, both balls now have vertical velocity V_vert1 = g * 1sec = 9.81 m/s² * 1 sec = 9.81 meters/second downward. (Initial vertical velocity was zero.)
At t = 1 second, Sandy's ball has a horizontal velocity V_hor = 30 m/s. Because there haven't been any forces with a horizontal component acting on it.
posted by sebastienbailard at 8:38 PM on January 22, 2016
It would be tidier to say, Sandy's ball has been accelerated horizontally and now has a non-zero horizontal velocity 'V_hor_initial = 30 m/s', while both balls still have zero vertical velocity.
Upon release, gravity accelerates both balls downward. After a time t, the vertical component of both balls' velocity is V_vert = g*t.
So at t = 1 second, both balls now have vertical velocity V_vert1 = g * 1sec = 9.81 m/s² * 1 sec = 9.81 meters/second downward. (Initial vertical velocity was zero.)
At t = 1 second, Sandy's ball has a horizontal velocity V_hor = 30 m/s. Because there haven't been any forces with a horizontal component acting on it.
posted by sebastienbailard at 8:38 PM on January 22, 2016
This means that x amount of newtons move it forward towards where it will strike the ground when gravity pulls it there. Will this not cause it to take a longer amount of time to hit the ground . . .
No. You can describe motion in terms of its movement in three dimensions. Those three dimensions are arbitrary, so for our purposes we can choose x being the direction the ball is thrown, y being perpendicular to x straight down to the ground, and z being perpendicular to x and y; because z doesn't change in this example, we can ignore it.
First, consider the dropped ball. Over the course of the experiment, the ball doesn't move in the x dimension. In the y dimension, the ball falls due to the acceleration due to gravity.
Now consider the thrown ball. Over the course of the experiment, the ball moves in the x dimension at the velocity v that Sandy throws it. Ignoring air resistance, and if it doesn't hit something, that ball will consider moving at v forever. Now consider the y dimension, and remember that x and y are separate. The ball isn't moving in the y dimension, only the x dimension, so in y, it starts falling due to the acceleration from gravity. How long does it take to hit the ground? As long as it takes to fall from its height, same as the dropped ball.
posted by Special Agent Dale Cooper at 8:39 PM on January 22, 2016
No. You can describe motion in terms of its movement in three dimensions. Those three dimensions are arbitrary, so for our purposes we can choose x being the direction the ball is thrown, y being perpendicular to x straight down to the ground, and z being perpendicular to x and y; because z doesn't change in this example, we can ignore it.
First, consider the dropped ball. Over the course of the experiment, the ball doesn't move in the x dimension. In the y dimension, the ball falls due to the acceleration due to gravity.
Now consider the thrown ball. Over the course of the experiment, the ball moves in the x dimension at the velocity v that Sandy throws it. Ignoring air resistance, and if it doesn't hit something, that ball will consider moving at v forever. Now consider the y dimension, and remember that x and y are separate. The ball isn't moving in the y dimension, only the x dimension, so in y, it starts falling due to the acceleration from gravity. How long does it take to hit the ground? As long as it takes to fall from its height, same as the dropped ball.
posted by Special Agent Dale Cooper at 8:39 PM on January 22, 2016
These Khan Academy videos on horizontally launched projectiles might be of interest.
posted by zachlipton at 8:46 PM on January 22, 2016
posted by zachlipton at 8:46 PM on January 22, 2016
Here's one way to think of it - if Sandy threw the ball straight down it would clearly hit the ground first, right? Also, if he threw it straight up in the air, then the dropped ball would hit first. Similarly, if Sandy threw the ball just a tiny bit down from level, wouldn't it make sense his ball would hit the ground first, as it's got a force pushing it down just a smidge more than the three year old's. And if Sandy threw it just a tiny bit up, it would seem to make sense it would stay in the air just a little bit longer than the other ball. So a perfectly level ball is where Sandy's ball matches the three year old's in time it takes to hit the ground.
posted by forforf at 8:47 PM on January 22, 2016 [6 favorites]
posted by forforf at 8:47 PM on January 22, 2016 [6 favorites]
Coming around to word-stuff:
They are of equal weight and mass.
Yup.
However Sandy's ball has been impelled from its inertia by the force of Sandy's throwing motion.
Not really. We would say, 'It experienced a force of yadda-yadda newtons for a while. Say 30 newtons, for half a second'. Its inertia, aka mass, aka obstinate desire to resist acceleration, didn't change. Its horizontal momentum m*V_hor increased from zero to m*30meters/s during the time it was being hurled.
This means that x amount of newtons move it forward towards where it will strike the ground when gravity pulls it there.
Would say, 'its horizontal velocity moves it forward', or 'its horizontal momentum moves it forward', (or 'its velocity moves it forward' or 'its momentum moves it forward').
After the moment of release, it is experiencing zero newtons of force in the horizontal direction. Which is why it isn't accelerating horizontally (or decelerating horizontally). Neglecting air friction, natch. During the half-second of hurling it was experiencing force, which we measure in newtons, and was for that reason accelerating.
posted by sebastienbailard at 8:50 PM on January 22, 2016
They are of equal weight and mass.
Yup.
However Sandy's ball has been impelled from its inertia by the force of Sandy's throwing motion.
Not really. We would say, 'It experienced a force of yadda-yadda newtons for a while. Say 30 newtons, for half a second'. Its inertia, aka mass, aka obstinate desire to resist acceleration, didn't change. Its horizontal momentum m*V_hor increased from zero to m*30meters/s during the time it was being hurled.
This means that x amount of newtons move it forward towards where it will strike the ground when gravity pulls it there.
Would say, 'its horizontal velocity moves it forward', or 'its horizontal momentum moves it forward', (or 'its velocity moves it forward' or 'its momentum moves it forward').
After the moment of release, it is experiencing zero newtons of force in the horizontal direction. Which is why it isn't accelerating horizontally (or decelerating horizontally). Neglecting air friction, natch. During the half-second of hurling it was experiencing force, which we measure in newtons, and was for that reason accelerating.
posted by sebastienbailard at 8:50 PM on January 22, 2016
Mefi's own Adam Savage did this on Mythbusters, if you are interested.
posted by It's Never Lurgi at 9:01 PM on January 22, 2016
posted by It's Never Lurgi at 9:01 PM on January 22, 2016
One of the hardest things to get your head around in two (or three) dimensional motion is that the dimensions remain segregated. Horizontal ("x") motion cannot affect vertical ("y") motion and y motion cannot affect x motion. It's not intuative, but it's true.
x and y live independent lives.
Much like Vegas, what happens in x stays in x.
posted by Betelgeuse at 9:07 PM on January 22, 2016 [9 favorites]
x and y live independent lives.
Much like Vegas, what happens in x stays in x.
posted by Betelgeuse at 9:07 PM on January 22, 2016 [9 favorites]
There's really no way to know, but they probably don't hit the ground at the same time.
That's because Koufax (and pretty much any decent MLB pitcher) puts spin on the ball when they pitch it. When we hear references to a "curveball" we think it's going left or right, but a lot of curve balls go down, and if not hit or caught, they'd hit the ground before the one the kid dropped. (If the pitcher puts forward spin on the ball.)
The reason for a "sinker" is that it fools the batter into swinging high, and even if he does hit it then it'll be a grounder and he'll probably be out.
On the other hand, it might have backspin and in that case it will not drop as fast as the batter expects; he might swing low, resulting in a hanging flyball which is also is an easy out.
Even if he isn't doing those things, a pitcher is throwing slightly downwards. He's standing on the mound, remember, and the strike box is below the point where he releases the ball, so its vertical motion isn't just a matter of gravity; it's got some downward velocity when he releases it.
posted by Chocolate Pickle at 9:23 PM on January 22, 2016
That's because Koufax (and pretty much any decent MLB pitcher) puts spin on the ball when they pitch it. When we hear references to a "curveball" we think it's going left or right, but a lot of curve balls go down, and if not hit or caught, they'd hit the ground before the one the kid dropped. (If the pitcher puts forward spin on the ball.)
The reason for a "sinker" is that it fools the batter into swinging high, and even if he does hit it then it'll be a grounder and he'll probably be out.
On the other hand, it might have backspin and in that case it will not drop as fast as the batter expects; he might swing low, resulting in a hanging flyball which is also is an easy out.
Even if he isn't doing those things, a pitcher is throwing slightly downwards. He's standing on the mound, remember, and the strike box is below the point where he releases the ball, so its vertical motion isn't just a matter of gravity; it's got some downward velocity when he releases it.
posted by Chocolate Pickle at 9:23 PM on January 22, 2016
I think it's possible your choice of example may be interfering with your intuitive understanding, since pitchers don't necessarily pitch balls perfectly horizontal. supercres's train example is a much better model for thinking of this since it demonstrates how the horizontal speed has no effect on the vertical drop. In fact, let's take it to an extreme — When you're on a plane and you drop a pen or a peanut or whatever, does it drift to the floor in a leisurely fashion, since you're traveling 600 mph horizontally?
posted by ejs at 9:29 PM on January 22, 2016
posted by ejs at 9:29 PM on January 22, 2016
If you account for the curvature of the Earth and the Earth's rotation, you'd find that Sandy's ball does not hit the ground at precisely the same moment, assuming its initial trajectory was tangent from the surface. But we're talking a very, very tiny difference.
posted by qxntpqbbbqxl at 9:33 PM on January 22, 2016 [1 favorite]
posted by qxntpqbbbqxl at 9:33 PM on January 22, 2016 [1 favorite]
And if you're wondering, air resistance in still air and with no spin will cause the thrown ball to fall slower (will experience greater y air resistance due to large x speed, because air resistance is nonlinear). But in the real world, ball spin and wind would cause much larger effects. There's also an effect from the sewn seams on the baseball, but I'm not sure exactly how that works.
posted by ryanrs at 9:49 PM on January 22, 2016
posted by ryanrs at 9:49 PM on January 22, 2016
How about this: if you're on a train going 90 mph and drop a ball from 5 feet off the floor, will it take the same time to hit the floor as one dropped from 5 feet while stationary? That's more obviously "yes", right?
This is really an excellent answer which not only makes what actually happens much more intuitive in a context of Newtonian dynamics, but also points the way to Einstein's theories of relativity since it's only one step from thinking about the moving railway car and what events inside it would look like to people standing on the ground outside and vice versa, to a more general consideration of different frames of reference and how the laws of physics in those frames of reference must relate as observers in each frame look at exactly the same events.
posted by jamjam at 9:51 PM on January 22, 2016 [2 favorites]
This is really an excellent answer which not only makes what actually happens much more intuitive in a context of Newtonian dynamics, but also points the way to Einstein's theories of relativity since it's only one step from thinking about the moving railway car and what events inside it would look like to people standing on the ground outside and vice versa, to a more general consideration of different frames of reference and how the laws of physics in those frames of reference must relate as observers in each frame look at exactly the same events.
posted by jamjam at 9:51 PM on January 22, 2016 [2 favorites]
How about this: if you're on a train going 90 mph and drop a ball from 5 feet off the floor, will it take the same time to hit the floor as one dropped from 5 feet while stationary? That's more obviously "yes", right?
And likewise you can also imagine that you are sitting on the train moving at 90 mph looking out the side window. The ball is thrown parallel to the train so that it is just outside the window moving exactly as fast as the train. From your point of the view on the moving train the ball doesn't seem to be moving horizontally at all. It remains right in front of your window, and the only motion you see is it dropping vertically to the ground, just like the stationary case.
posted by JackFlash at 10:21 PM on January 22, 2016 [3 favorites]
And likewise you can also imagine that you are sitting on the train moving at 90 mph looking out the side window. The ball is thrown parallel to the train so that it is just outside the window moving exactly as fast as the train. From your point of the view on the moving train the ball doesn't seem to be moving horizontally at all. It remains right in front of your window, and the only motion you see is it dropping vertically to the ground, just like the stationary case.
posted by JackFlash at 10:21 PM on January 22, 2016 [3 favorites]
The simple equation of motion for a body under uniform acceleration is all that is needed to demonstrate this phenomenon.
u = 0.5*a*t^2 + Vo*t + Xo
(where u is the displacement of the object, a is the constant acceleration, t is time, Xo is the initial position, and Vo is inital velocity)
This problem is in three dimensions so let's vectorize this equation.
[ux uy uz] = 0.5*[ax ay az]*t^2 + [Vox Voy Voz]*t + [Xox Xoy Xoz]
(this equation represents 3 separate equations, one for each dimension of space. For example, the displacement in the x direction is given as ux = 0.5*ax*t^2 + Vox*t + Xox )
Taken from the moment the ball leaves the pitcher's hand and the boy's hand, we know the acceleration vector and initial velocity vector for both the pitch and the ball that is dropped. Let us say that the pitch speed is 50 meters per second horizontally.
The pitch: a = [0 -9.8 0]; Vo = [50 0 0]; Xo = [25 5 0]
The dropped ball: a = [0 -9.8 0]; Vo = [0 0 0]; Xo = [0 5 0]
Apply both sets of vectors in the equation for u above and you will see that the component uy is identical for all values of time t for both the pitched and dropped ball. Also, notice how this equation gives this result and there is no mention anywhere of mass or weight.
posted by incolorinred at 10:22 PM on January 22, 2016
u = 0.5*a*t^2 + Vo*t + Xo
(where u is the displacement of the object, a is the constant acceleration, t is time, Xo is the initial position, and Vo is inital velocity)
This problem is in three dimensions so let's vectorize this equation.
[ux uy uz] = 0.5*[ax ay az]*t^2 + [Vox Voy Voz]*t + [Xox Xoy Xoz]
(this equation represents 3 separate equations, one for each dimension of space. For example, the displacement in the x direction is given as ux = 0.5*ax*t^2 + Vox*t + Xox )
Taken from the moment the ball leaves the pitcher's hand and the boy's hand, we know the acceleration vector and initial velocity vector for both the pitch and the ball that is dropped. Let us say that the pitch speed is 50 meters per second horizontally.
The pitch: a = [0 -9.8 0]; Vo = [50 0 0]; Xo = [25 5 0]
The dropped ball: a = [0 -9.8 0]; Vo = [0 0 0]; Xo = [0 5 0]
Apply both sets of vectors in the equation for u above and you will see that the component uy is identical for all values of time t for both the pitched and dropped ball. Also, notice how this equation gives this result and there is no mention anywhere of mass or weight.
posted by incolorinred at 10:22 PM on January 22, 2016
In a simple model of this problem, there are two axes, x and y. The "throwing motion" is entirely on the x axis ("forward"). The force of gravity is entirely on the y axis ("down"). Our only question is, "when does the ball reach position y1 on the y axis (the ground)?"
Let's look at the ball's position now, a moment after it is released. It is moving forward (in the x direction) because we gave it an initial velocity with the throw. It is also being accelerated in the y direction by the force of gravity. We only care about the position in the y direction for the purposes of our question.
Now picture the non-thrown ball. It is also being accelerated by gravity in the y direction at the same rate. I does not, however, have any initial velocity in the x direction, because we didn't throw it.
Does the lack of forward motion (x axis) make a bit of difference in the problem? Nope. Our problem looks exactly the same no matter if the ball is moving forward (on the x axis) at 1,000 mph or 0 mph, since we only care about the ball's position above the ground (on the y axis).
posted by deathpanels at 10:24 PM on January 22, 2016
Let's look at the ball's position now, a moment after it is released. It is moving forward (in the x direction) because we gave it an initial velocity with the throw. It is also being accelerated in the y direction by the force of gravity. We only care about the position in the y direction for the purposes of our question.
Now picture the non-thrown ball. It is also being accelerated by gravity in the y direction at the same rate. I does not, however, have any initial velocity in the x direction, because we didn't throw it.
Does the lack of forward motion (x axis) make a bit of difference in the problem? Nope. Our problem looks exactly the same no matter if the ball is moving forward (on the x axis) at 1,000 mph or 0 mph, since we only care about the ball's position above the ground (on the y axis).
posted by deathpanels at 10:24 PM on January 22, 2016
ryanrs: "And if you're wondering, air resistance in still air and with no spin will cause the thrown ball to fall slower (will experience greater y air resistance due to large x speed, because air resistance is nonlinear)."
This doesn't seem right; at least for an object that can be considered uniform for the purpose of the thought experiment posed. All the velocity induced drag would be operating directly opposite the forward motion and lie on the horizontal axis. While the vertical movement will result in a slight off horizontal vector to the air resistance the vertical component of the vector would be the same for both balls.
posted by Mitheral at 11:40 PM on January 22, 2016
This doesn't seem right; at least for an object that can be considered uniform for the purpose of the thought experiment posed. All the velocity induced drag would be operating directly opposite the forward motion and lie on the horizontal axis. While the vertical movement will result in a slight off horizontal vector to the air resistance the vertical component of the vector would be the same for both balls.
posted by Mitheral at 11:40 PM on January 22, 2016
FYI the classic example problem of this, often given in freshman physics courses, is called the "monkey and hunter" problem or the "monkey shooting" problem. You can easily find various explanations of it, as well as video demos, if you search for those terms.
posted by kickingtheground at 12:15 AM on January 23, 2016
posted by kickingtheground at 12:15 AM on January 23, 2016
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How about this: if you're on a train going 90 mph and drop a ball from 5 feet off the floor, will it take the same time to hit the floor as one dropped from 5 feet while stationary? That's more obviously "yes", right? Same trajectory as your setup, again, assuming a perfectly level 90 mph throw.
posted by supercres at 8:36 PM on January 22, 2016 [11 favorites]