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 = I*W.*

Momentum of bat= MV = MWL

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

+(1/24)(W^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.