the physics of throwing a baseball

[asofel]

Some friends of mine had an interesting debate at a party this weekend. The issue was this: when a pitcher throws a baseball, as soon as its released from his hand, does it accelerate and pickup any speed, or does it start slowing down immediately due to air resistance. I'm curious as to what others think.

[Patrick del Poker Grande]

Technically, it's essentially always accelerating. In the context of the way you and your friends are talking, though, it's slowing down as soon as it leaves the pitcher's hand.

[2planka]

The "muzzle velocity" of a pitched baseball slows down about 1 mph every 7 feet after it leaves the pitcher's hand, that's a loss of roughly 8 mph by the time it crosses the plate.

From Steve Orinick's Baseball Physics page at his site. (http://www.stevetheump.com/baseball_umpire_frames.htm)

[ripdog]

[ QUOTE ]
The "muzzle velocity" of a pitched baseball slows down about 1 mph every 7 feet after it leaves the pitcher's hand, that's a loss of roughly 8 mph by the time it crosses the plate.

From Steve Orinick's Baseball Physics page at his site. (http://www.stevetheump.com/baseball_umpire_frames.htm)

[/ QUOTE ]

That loss of velocity doesn't seem to match the equation from my physics book--it's way too low. The equation is:

a=V^2/2X

a is arm acceleration in m/s
V is velocity of pitch in m/s
X is the distance that the ball is accelerated in meters

So if you want to throw a ball at 85mph (~38 m/s) and have an X of 11.375' (3.5 meters), your arm must accerate the ball to ~206 m/s or ~461mph. If we assume that you let the ball go at a distance of 50' from the plate, the ball loses an average of ~7.5mph per foot.

[mason55]

[ QUOTE ]
[ QUOTE ]
The "muzzle velocity" of a pitched baseball slows down about 1 mph every 7 feet after it leaves the pitcher's hand, that's a loss of roughly 8 mph by the time it crosses the plate.

From Steve Orinick's Baseball Physics page at his site. (http://www.stevetheump.com/baseball_umpire_frames.htm)

[/ QUOTE ]

That loss of velocity doesn't seem to match the equation from my physics book--it's way too low. The equation is:

a=V^2/2X

a is arm acceleration in m/s
V is velocity of pitch in m/s
X is the distance that the ball is accelerated in meters

So if you want to throw a ball at 85mph (~38 m/s) and have an X of 11.375' (3.5 meters), your arm must accerate the ball to ~206 m/s or ~461mph. If we assume that you let the ball go at a distance of 50' from the plate, the ball loses an average of ~7.5mph per foot.

[/ QUOTE ]

That's pure linear acceleration. Pitchers use centripetal acceleration.

*waits for Patrick to come correct me*

[Patrick del Poker Grande]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
The "muzzle velocity" of a pitched baseball slows down about 1 mph every 7 feet after it leaves the pitcher's hand, that's a loss of roughly 8 mph by the time it crosses the plate.

From Steve Orinick's Baseball Physics page at his site. (http://www.stevetheump.com/baseball_umpire_frames.htm)

[/ QUOTE ]

That loss of velocity doesn't seem to match the equation from my physics book--it's way too low. The equation is:

a=V^2/2X

a is arm acceleration in m/s
V is velocity of pitch in m/s
X is the distance that the ball is accelerated in meters

So if you want to throw a ball at 85mph (~38 m/s) and have an X of 11.375' (3.5 meters), your arm must accerate the ball to ~206 m/s or ~461mph. If we assume that you let the ball go at a distance of 50' from the plate, the ball loses an average of ~7.5mph per foot.

[/ QUOTE ]

That's pure linear acceleration. Pitchers use centripetal acceleration.

*waits for Patrick to come correct me*

[/ QUOTE ]
Well, I didn't care to really look at the equations posted, but they don't look right at first glance (literally about half a second). Also, it is linear acceleration we're interested in once the ball leaves the pitcher's hand.

[Voltron87]

I dont know too much about physics relative to people who know what theyre talking about on this board, but I cant see a way the ball would speed up. from a common sense point of view, once the ball leaves the pitchers hand there is no more force applied on it in the direction towards the plate, but there is air resistance, which slows it down. if thats the only force, it will slow down.

[mason55]

[ QUOTE ]

Well, I didn't care to really look at the equations posted, but they don't look right at first glance (literally about half a second). Also, it is linear acceleration we're interested in once the ball leaves the pitcher's hand.

[/ QUOTE ]

Right, but I got the impression that he was talking about the pitcher accelerating the ball through the wind up, saying that it would have be accelerated to 460mph. I'm still probably wrong though, but those equations definitely aren't right.

[Victor]

[ QUOTE ]
I dont know too much about physics relative to people who know what theyre talking about on this board, but I cant see a way the ball would speed up. from a common sense point of view, once the ball leaves the pitchers hand there is no more force applied on it in the direction towards the plate, but there is air resistance, which slows it down. if thats the only force, it will slow down.

[/ QUOTE ]

acceleration, in an academic physics sense, does not necessarily mean that it will speed up. acceleration is simply change in velocity, which the is always occuring with a pitched ball (i.e. slowing down is acceleration).

[brassnuts]

It will immediately begin to slow down in the horizontal direction due to wind resistance. At the same time, it will begin to speed up in a downward direction due to gravity. If he acceleration in the the downward direction is greater than the acceleration due to wind resistance, then the ball is speeding up, and vise versa. I'm not sure which acceleration is greater, but my intuition says the force of the wind resitance is greater than the force of gravity, so I'm gonna say that it slows down immediately after being released. Actually, if someone had the coefficients handy, I'm sure it could very easily be calculated.

[goofball]

an interesting side note:

the reasons balls break near the plate: they are going slower, so the forces acting on them in the y and z planes (we consider x to be from batter to center field) appear mroe pronounced. This isn't because they are, but because the v(x) is slower, the v(y,z) (cause be a in those same directinos caused by gravity and wind resistance turbulence) appears to be higher. It's only higher relative to v(x) but as such, the ball moves a greater distance in the y,z planes for every unit distance it moves in the x plane.

[brassnuts]

[ QUOTE ]
It will immediately begin to slow down in the horizontal direction due to wind resistance. At the same time, it will begin to speed up in a downward direction due to gravity. If he acceleration in the the downward direction is greater than the acceleration due to wind resistance, then the ball is speeding up, and vise versa. I'm not sure which acceleration is greater, but my intuition says the force of the wind resitance is greater than the force of gravity, so I'm gonna say that it slows down immediately after being released. Actually, if someone had the coefficients handy, I'm sure it could very easily be calculated.

[/ QUOTE ]

And this is wrong. I forgot that we're dealing with vectors here, so it's not as simple as "If he acceleration in the the downward direction is greater than the acceleration due to wind resistance, then the ball is speeding up, and vise versa." Because of this, the ball is undoubtably slowing down as soon as it leaves the hand.