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Physics & Baseball

  1. Jun 23, 2010 #1
    Recently a friend and I were discussing a video on MLB.com about a 5 year old hitting a 90MPH fastball. I said no way would even have the arm strength to stop the ball let alone hit the ball.

    That started us wondering......... how much force is actually needed to make a 90MPH fast ball going in one direction, stop it, and send it careening in a totally new direction at say 60MPH? I bet a line driive has to be going at least that fast.

    So how bout it guys, anyone have or can owrk out an equation for that? Would love to really see how much force is required to hit off a MLB Pitcher!!!
  2. jcsd
  3. Jun 23, 2010 #2


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    Depending on the weight of the bat, just holding the bat still would result in the baseball bouncing off at some speed. And it wouldn't take much speed of the bat to have the ball come off at 60mph with a 90mph pitch.
  4. Jun 23, 2010 #3


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    Very little at all, as long as the baseball hits the bat near its "sweet spot", if the kid can swing the bat at all, he should have no problem sending the baseball at 60mph. Probably a lot faster. What's the regulation weight for bat and the baseball? I know bats have ranges, but maybe typical? We can make some estimates then.
  5. Jun 23, 2010 #4
    MLB Baseballs must weight between 5-5.25 ounces

    Most MLB bats weigh between 30-33 ounces. All are -3 drop, meaning if it is a 35 inch long bat, it must be 32 ounces. Though TECHNICALLY it can be as long as you want as long as its -3 drop most keep them 33-38 inches in length

    So lets say the ball is 5.125 ounces, the bat is 35 inches and weighs 32 ounces.

    The mound is is 60'6" from home plate, MLB Radar Guns typically hit the ball about 5-10' in front of home plate for the MPH number. Figure a good line drive then goes at least 150 feet and keeps its momentium up around 60-70 MPH (if not faster) till it hits the ground.

    Just how much force is Big Papi hitting that sucker with?

    Ya know i love that the internet can turn a conversation between 2 friends over to smart guys like you who can actually provide numbers on somethgin like this!!!!!
  6. Jun 23, 2010 #5


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    I just realized I have no idea how elastic that thing is, and don't have any idea how to estimate it without an actual baseball...

    For a moment, pretend that it was a 5 ounce superball hitting a 30 ounce bat at 90mph. Lets take the bat as stationary. The combined center of mass is moving at 90/(30+5) = 2.57mph, and it will keep going at that speed after collision (conservation of momentum). The "baseball" is then catching up to the center of mass at 87.43mph, until it catches up with it in collision. IF this was a perfectly elastic collision, it would go back at the same 87.43mph relative to the center of mass, which would put final speed at 84.86 mph, with kid having to do nothing more than hold the bat up in the right spot.

    Of course, that feels wrong for a baseball. If you threw a baseball at 87.43mph into a solid wall, it wouldn't bounce back at 87.43mph. It'd still bounce off quite fast, but not that fast.

    Any idea how fast that would be? That would tell us the final speed of the baseball. Otherwise, I need to track down a baseball and do science to it. Not sure I know anyone who plays it...

    60mph is still very reasonable at a relatively mild swing, but I need more data to say how mild of a swing. Maybe if I can find some statistics on hits that have been made, like, speed in/out and the speed of the bat. I'd then be able to extrapolate everything I need.
  7. Jun 24, 2010 #6


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    We can estimate this without having to know the force. Conservation of momentum, and knowing the coefficient of restitution, is all that is needed. Let's see how it works out for a stationary bat.

    The coefficient of restitution is the ratio of the relative velocity of bat and ball after and before the collision:
    c.o.r. = |vbat - vball| / 90 mph​
    where vball and vbat are the velocities after the collision.

    A reasonable value for the c.o.r. seems to be 0.5 to 0.6, according to two online sources I found***. Using 0.55, and the fact that momentum is conserved:
    c.o.r.: 0.55 = (vbat - vball) / 90 mph

    momentum: 5.1*90 mph = 5.1*vball + 31*vbat
    I'm taking the initial 90 mph to be in the positive direction. As long as the ball bounces backwards off the stationary bat -- i.e. vball is negative -- we can consider it a successful hit.

    Solving the above equations gives -30 mph for the ball's velocity. So yes, anybody who can hold a bat has the strength to "hit" a ball -- as long as the bat and ball make good contact.

    Hope that helps.


    *** coefficient of restitution for a bat and ball:
    see section 2 http://www.physics.usyd.edu.au/~cross/baseball.html" [Broken]
    see figure 5 http://www.docstoc.com/docs/20726020/The-Coefficient-of-Restitution-of-Baseballs-as-a-Function" [Broken]
    Last edited by a moderator: May 4, 2017
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