Physics question (slightly off-topic)

a better place to ask this question

http://physics.stackexchange.com/

The one which goes at 90 mph. For details see this subcomment.

Now I wonder what it is like to be a bat.

Meta-comment: I know that the question is trivial and this isn't a physics discussion board, but still am surprised by the number of downvotes. I am interested in the voters' reasons. Is it OK to ask for an advice concerning diet, headaches or car purchase, but not physics? Why is that?

Notice: The post has been at -5 at one moment, the comment reflects that state. If its long-term karma is positive or zero, the implicit hypothesis in my question - i.e. that people are downvoting questions about physics more than questions about headaches - may not be true.

Well, if we ignore the force being exerted on the bat by the batter, then we can use the principle of conservation of momentum, and if we make-believe that the collision is perfectly elastic, then we can say that kinetic energy is conserved as well. This gives us two equations to play with:

where the vs are initial velocities, ms are masses, and ws are subsequent (after-collision) velocities. Algebra then lets us derive two equations for the subsequent velocity of the ball in terms of the masses of the bat and ball, the subsequent and initial velocities of the bat, and the initial velocity of the ball:

and

In both these equations [edit: this is wrong; see addendum below], the subsequent velocity of the ball is an increasing function of the initial velocity of the ball (when all the other variables are treated as constants). Clearly, balls that are hit faster will travel farther, so this would seem to suggest that the faster pitch will be hit farther. However, in addition to the simplifying assumptions we made at the beginning, this analysis also assumes that the after-collision velocity of the bat is independent of the speed of the pitch, which is surely not true, so this reasoning cannot be said to have definitively resolved the question.

EDITED TO ADD: Except velocity and momentum are vector quantities, so the initial velocity of the ball should be regarded as having sign opposite of that of the initial velocity of the bat. This would seem to mean that our momentum equation is actually saying that the slower pitch will be hit further ... I'm confused.

(%i1) solve([m*v+mm*vv=m*w+mm*ww, m*v*v+mm*vv*vv=m*w*w+mm*ww*ww], [w,ww]);
(%o1) [[w=v,ww=vv],[w=(2*mm*vv+(m-mm)*v)/(mm+m),ww=((mm-m)*vv+2*m*v)/(mm+m)]]

The problem is complicated, so lets simplify and then look at two limiting cases and see what they suggest.

First, it is simpler to ask which ball leaves the bat in a "forward" direction faster. Presumably the one that leaves faster also goes further. The collision is "elastic" neither the bat nor the ball gets permananetly crushed in a way as to take energy out of the system.

The two limiting cases to look at 1) bat is moving at 0 velocit and weighs same as ball Then either the 60 or the 90 mph ball come to a dead stop on contact with the bat and the bat recedes at 60 or 90 mph backwards. 2) the bat is moving at 0 velocity and weighs a tremendous amount more than the ball. Then the (60, 90) mph ball hitting the very heavy stationary "bat" bounce off it in the forward direction at (60, 90) mph and the bat remains essentially stationary.

So we see for bats between the weight of the ball and infinity (and I think this is the true range of bat weights), the 60 mph ball never goes faster than the 90 mph ball, but sometimes the 90 mph ball goes faster. I think details of how fast the bat is actually going will not matter, that we have surrounded the problem with this simplification, and the fast ball goes further when hit well.

Assuming in both cases the ball is struck squarely, the one going 90 mph, though note that the speed of the ball affects the probability that the ball is struck squarely (or at all).

Most likely, the one at 90. The ball bounces, so moving it towards the bat faster makes it faster when it bounces off.

Others have answered in equations; let me see if I can put the answer in words, for possibly easier understanding. When ball hits bat, its kinetic energy is (with some reasonably high efficiency) converted into potential energy by the ball deforming from its equilibrium shape; then that gets reconverted to kinetic energy (now going in the opposite direction) as the ball springs back to roundness. In other words, the ball bounces. (If you assume a rigid ball the problem changes a lot.) So we can rephrase the question: Which will bounce higher, a ball that's just dropped from some given height, or a ball that is thrown downward so it hits the ground at a higher speed? Now we intuitively see that it is the latter. Bouncing from a moving bat is no different from bouncing off the ground.

Now there are some caveats to that; the efficiency of the energy conversion is not constant, it's some function of the ball's speed - at some point the ball doesn't bounce at all, it just disintegrates (or bursts into flames, if the loss to heat is sufficient). But for speeds achievable by human throwing arms, this effect can probably be ignored.

Generic intuition pump: take much more extreme values: what happens if the ball is thrown at a million miles per hour? Maybe it bounces back at a million miles per hour, maybe the batter looks in dismay at the pile of splinters that used to be his bat ... so it depends of what your model of the batter is. The question as asked doesn't allow to answer that question, though you could probably estimate a reasonable model of a batter.