It seems that major league baseball players have to wait a relatively long time to field a pop-up. This led me to wonder how high the ball traveled. If one knew how high the ball traveled, one could calculate speed and time approximately.


Not sure if you're looking for an "average" popup or exceptional ones.

A website with some facts on Tropicana Field says the height of the "A" ring catwalks ranges from 181 feet in center to 194 feet near home plate. A Cleveland.com piece has similar figures, and also suggests that as of 2008 (representing play over 18 seasons), no one had hit the catwalks at that height.

A New York Times article states that two players had hit the "A" ring as of 2010, including Jason Kubel at "about 190 feet" high. Balls could (rarely) be hit higher than this in other locations and wouldn't be noted. So I'd suggest an upper limit to flies at around 200 feet.

  • This is kind of info I'm looking for. I was looking for an average range of values but max value is fine and really is more interesting. I read somewhere that highest pop-up was about 177 feet. Close to your number as well as the numbers you gave from the Times. I guess fangraph has the numbers for past few years but never mentions them.
    – Natsfan
    Jun 18 '18 at 17:06

Joey Gallo hit a pop up on April 21, 2019 against the Houston Astros that was measured by StatCast to have reached 207 feet at its apex and had a hang time of 7.3 seconds.

Off the bat, the shortstop headed to short right field, but the wind brought the ball back to the infield and it dropped at the shortstop position for an infield base hit.


So, if we make a few assumptions, we can get a reasonable estimate.

First, I'm going to take the time from contact to when it's fielded at 6 seconds -- I got this by taking the video of Hosmer dropping the Bregman popup, which took 24 seconds at 1/4 playback speed; i.e., 6 seconds of real time.

Next, I'm going to disregard air resistance to simplify the physics. From here, we can invoke two elementary physics formulas:

  • t = (2*v*sin(theta)) / g
  • h = (v^2*sin(theta)^2) / (2*g)

where v is the exit velocity off the bat, theta is the launch angle, and g is the gravitational acceleration (9.8 m/s^2).

Rearranging the first equation for v gives v = g*t / (2*sin(theta)). Now, we can plug that into the second equation. Doing so (after some algebra) gives h = g*t^2 / 8. Very conveniently, we don't need to know (estimate) the exit velocity or the launch angle. Now, we simply insert our values, which gives h = (9.8 m/s^2) * (6 s)^2 / 8 = 44.1 m, which is roughly 145 feet. Hence, we can conclude that a popup can go roughly 145 feet high.

  • 2 points. You calculate height but you use the total time for trip up and down. Also this does not consider atmospheric drag which is an important term, and makes the problem much harder. I could do the calculation but I'm unsure about the atmospheric drag. I was hoping someone knew of an actual measured value. Interestingly, your answer is close to what I expected though. Thanks for input.
    – Natsfan
    Jun 17 '18 at 23:48
  • On the first point, the time equation is for the total time that the object is in the air, not just the trip up or the trip down. So, the proper time to use here is the 6 second estimate. On the second, the drag is certainly important, but using that makes the physics much worse quickly. I would suspect that if we included it, that the height would get larger, because the speed of the ball will decrease more than it would if it travelled in a vacuum, but the time will remain constant. So, we can take the 145 feet as a lower bound. Jun 18 '18 at 14:29
  • I'll have to work thru this myself. If the 6 seconds is correct, that is the time to go up and back down. So it shouldn't give you the apex of the flight unless you took half the distance traveled.. Also, the drag will reduce the height because the ball will reach zero speed quicker. At a lower height. Doesn't make sense for a drag force, a force opposing the motion, to increase the distance traveled.
    – Natsfan
    Jun 20 '18 at 22:45
  • Yes, you're right about the change in drag. I'd flipped the distance = velocity*time equation in my head. However, about the first part, the equations are what I claim they are. The h in the second equation is the apex height, and we can substitute the time equation into the height equation and find the height as a function of time. Now, it's certainly possible that there are other popups that have a flight time of more than 6 seconds. Jun 21 '18 at 1:35
  • thanks for the response. The confusion is probably on my part. I'll work thru it tomorrow and hopefully we'll agree.
    – Natsfan
    Jun 21 '18 at 2:01

Mike Laga hit a pop fly out of the old Busch Stadium.


Estimated at around 250 feet high and over 300 in the air to make it completely over the roof. Not sure I have ever seen a ball hit so high in my life.

  • Wow, what a shot! I assume it was caught since most fields have home runs at well over 300 ft.
    – Natsfan
    Jun 26 '18 at 14:28

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