(This might be more appropriate for the Game forum eventually, but I’m looking for a factual answer).
A few days ago, with his team down 12-3 to the Oakland Athletics in the bottom of the 8th, Rangers third baseman Brock Holt was called upon to make his third career relief pitching appearance.
His first pitch was recorded at 31 miles per hour for a strike: A high, arcing pitch that somehow made it back into the strike zone. The announcers assured us that pulling this off is much, much harder than it looks. According to other articles I’ve read, this is the slowest recorded pitch for a strike in the history of Major League Baseball.
You should watch the entire inning; it has a hilarious result.
How slow is it possible to throw a pitch from the mound and have it still land in the strike zone?
Trajectory calculator.
A launch at 8 ft. in height at 0 degrees angle requires a speed of 58.5 mph to reach the plate at ground level.
Hard to tell with the scale but likely 62 -63 to arrive at knee height(WAG).
The pitch in question is being thrown at a high arc so it’s traveling considerably farther than 60.5 ft.
If the number is that high, that would suggest that the velocity of the pitch is measure later, perhaps as it actually reaches the strike zone. Or it’s an average velocity of the whole thing.
Also I’m pretty sure the radar guns only measure the horizontal component of the velocity. In the absence of air friction, and given a powerful enough arm, it should be possible to make the horizontal component of the velocity arbitrarily small when it crosses the plate.
You can get more range with less speed, by throwing upwards at an angle. If you throw at 45 degrees, neglect air resistance, and assume that your starting height is the same as your final height, then your range will be R = v^2/g. From the pitcher’s mound to home plate is about 18.4 m, which would give a velocity of about 13.4m/s (30 MPH). If the radar gun is only measuring horizontal component, then divide that by sqrt(2), for 9.51 m/s (21.3 mph).
You could get away with a little bit less than that, if you release the ball at the top of your overhead reach and just skim the bottom of the strike zone, but it won’t make all that much difference, and it’s harder to calculate.
The effects of air resistance probably would require you to throw the ball a bit faster than these numbers, but they’re probably still in the right ballpark (so to speak.) Including air resistance is, of course, even harder than calculating the range when the release and target are not level with each other.
If you throw a ball up, it comes down - in a parabolic arc. Ignoring wind resistance (sort of like spherical chickens in a vacuum), if you throw an arc to a height of h then the vertical component of velocity V of the ball coming down to the same height is:
h=v^2/(2g) or v=sqrt(2gh)
Meanwhile, you are throwing with a horizontal component U for d=ut
h- height thrown,
v - vertical component of the pitch speed
u - horizontal component of pitch speed
d - distance to plate 18.4m
g - gravity, 9.81 m/s/s
t - time in the air
These tie together with
v=gt/2 (t/2 because the ball spends half the time of the pitch going up, and half coming down)
I reduce this with my 50-years-ago math to:
h=1660/(u^2)
The actual velocity of the pitch, if you are not just measuring U, horizontal is:
velocity = sqrt(v^2 + u^2)
After all, the higher you throw the ball, the less U has to be. The lower limit on U is how high one could throw a ball. (Max V)
If I can still do math…
If I throw upward at v=30m/s (98ft/sec or 67mph)
it will take 3.058 seconds to reach the top, at 45.9m (148ft) and another 3.058 sec to come back down
To traverse the distance to the plate, it only needs to be doing 0.33m/sec horizontally, or .75mph.
But its total velocity will be pretty close to the 30m/s vertical.
Radar guns only measure the component of the velocity along the line from the gun to the ball – radial velocity, that is. If the gun and ball are at the same height, this is the same thing as horizontal velocity. For small height differences it is very close, because the cosine of small angles is very close to 1. In a pathological case such as a high parabolic trajectory, it could be very different from horizontal. If you put the gun under the strike zone looking up, it’d have little resemblance to a horizontal velocity.
Right, when people are saying “radar guns measure the horizontal component”, what they mean is “radar guns, as they’re currently situated in real ballparks to measure ordinary pitches, measure the horizontal component”.
