Baseball: Hit farther off a tee, or hit from a pitch?

Something that I’ve been wondering about is whether a baseball player could hit a ball farther if he was hitting it from a tee vs. hitting it off a pitch. Physics-wise, it seems to me that he could hit it farther off a tee,
since he doesn’t have to overcome the velocity of the pitched ball to get it going the other way. But everyone who ‘knows’ about baseball says there is no question that a batter can hit it farther off a “fastball down the middle” than he ever could off a tee. At any rate, it doesn’t make any sense to me.

Anybody know?

If you hit a baseball off a tee, the batter has to supply all of the power.

With a pitched ball, you do have to overcome the velocity of the ball coming toward you, but the bat should be able to deliver enough energy to the ball to reverse its direction quite easily and even add velocity to it.

Why do you think that pitchers get hit in the head sometimes by line drives? The ball is coming back faster than they threw it.

Someone with a physics background can probably explain this a whole lot better.

I don’t have a physics background, but here’s how I could explain it.

You know those metal balls on wires that you see on people’s desks? You let one ball bounce and then the energy is transferred across the balls to the ball at the end and it moves out and then back? That same energy is in the thrown pitch. When a thrown ball hits a wall, the energy is transferred to the wall, which does not move, but transfers the energy straight back to the ball, but in the opposite direction. It’s the same with the bat. When the thrown ball hits the bat, the bat transfers the energy back to the ball in the opposite direction, plus the additional energy supplied by the batter by swinging the bat. The result is the ball now travels faster (assuming it wasn’t a bunt) in the opposite direction. If the batter hits a stationary ball off a tee, then there is no kinetic energy of the ball to be transferred in the opposite direction, only the potential energy that comes from the swinging bat.

Does that make sense?

Yeah, that’s how it works.

Some of the energy from the ball and bat is lost in the collision however as a bat is not perfectly elastic. If you could take a picture of the exact moment the ball and bat collide, you would see both of them deformed to a certain extent.

Not to mention the energy that is absorbed by the batter. If you’ve ever hit a fastball down the middle and felt that jolt through your arms and shoulders, you know what I’m talking about.

To take the extreme example, imagine a bunt. If you just hold out your bat and let it get hit with a 90 mile an hour fastball, it’s going to bounce off pretty far. A ball on a tee obviously won’t move at all. Now, consider a swing to be just a bunt+some movement on the bat.

There’s two things that you need to conserve here, energy (1/2 mv[sup]2[/sup], no direction) and momentum (mv, in the direction of motion). Any change in the ball must be countered by a change in the bat/batter/ground. The fastball must change its momentum more, but the teeball must change its energy more. Because velocity is squared in the energy formula, it’s harder to change the energy of the ball, so you get the most out of the hit by changing the energy a small amount and the momentum a large amount in a fastball.

Assume that the pitch and hit are at the same height, the deformation of the bat and ball is elastic and temporary (no plastic deformation) and that no heat is lost at any time. These are incredibly nit-picky points you can handedly ignore.

Say that the bat has a mass of M and the ball has a mass of m. The pitch has velocity +v where positive velocity is from the mound to the plate. The ball is radius r. Angular velocity is represented by w; moment of inertia by I. Density by p, gravitational constant by g, height by h.

Assume the pitch is not spinning, i.e. the angular velocity of the pitch is 0.

Then after being pitched, the ball has:
Momentum = mass*velocity = m(+v) = mv

Energy = kinetic + potential + rotational + elastic
=0.5 (mv^2 + 2pgh + Iw^2 + kx^2)

For a rod, Izz=(1/12)mr^2. For a sphere, Izz=mr^2.
We are ignoring potential energy since the height is assumed not to change. We will also ignore elastic potential energy.)

Energy (of ball) = 0.5 m(+v)^2 + 0 + 0.5 (mr^2)(0^2)
= 0.5m(v^2)

Momentum of bat = mass*velocity

Assume the bat is being swung with an angular speed of W. If the bat has a length of L, this corresponds to a velocity of (WL).

Angular Momentum of bat = IW.
Momentum of bat= M
Energy of bat = 0.5M(V^2)+0.5(1/12)(W^2)

If the ball was initially on a tee, all the energy would come from the swing, since in this case v=0.

In any case, energy is conserved (along with momentum and angular momentum) after the bat has stopped but the ball is still flying. Here, if all the energy of the bat is transferred to the ball

Energy (ball, bat stopped) = 0.5m(v^2) + 0.5M(WL)^2

Energy(pitch) - Energy (tee) = 0.5M(V)^2 + (1/24)(W^2)

The net velocity of the ball, v=(2E/m)^0.5

In essence, the system has more energy and the ball has higher velocity if it is pitched, not hit off a tee. It will go further. It is easy to make this more confusing by adding in air drag, etc. But I’ve probably told you too much already.