The fastest recorded pitch in major-league baseball, thrown by Nolan Ryan in 1974, was clocked at 100.8 mi/h. If a pitch were thrown horizontally with this velocity, how far would the ball fall vertically by the time it reached home plate, 60.0 ft away?
Easier explained,
Velocity = (162,300/ 3600s in 1 hr) = 45.08m/sec.
Time to travel 18.3m = (d/v), = 18.3/45.08) = .406 secs.
Height of fall in .406s. = 1/2 (t^2 x g), = .808m.
u = 162.3 km/h = 45.083 m/s
x = xo + u*t + 1/2 at^2
45.083*t + 0 = 18.3
t = 0.406 s
y = yo + u*t + 1/2 at^2
y + 0 - 4.9t^2 = 0
y = 4.9(0.406)^2
y = .808 m
the question was in English units ... not metric ... 2.64 ft
Well, let's do some physics math, shall we? If the pitch traveled horizontally at 100.8 mi/h, it didn't have time to think about falling vertically. Poor little ball didn't even have a chance to experience the thrill of flying and falling at the same time. It was way too focused on reaching home plate.
So, in short, the ball wouldn't fall at all vertically by the time it reached home plate. It would just be zooming through the air, begging the batter to give it a good hit.
To find how far the ball would fall vertically, we need to calculate the time it takes for the ball to reach home plate. We will assume that air resistance is negligible.
First, let's convert the velocity from miles per hour to feet per second:
100.8 mi/h * 1.47 ft/s per mi/h = 148.176 ft/s
We can use the equation of motion: d = (1/2) * g * t^2 to find the vertical distance fallen, where d is the vertical distance, g is the acceleration due to gravity, and t is the time of flight.
The acceleration due to gravity, g, is approximately 32.2 ft/s^2.
We know that the horizontal distance, x, traveled by the ball is 60.0 ft. The time of flight, t, can be calculated using the horizontal distance and horizontal velocity:
t = x / velocity
t = 60.0 ft / 148.176 ft/s ≈ 0.4048 s
Now we can calculate the vertical distance fallen:
d = (1/2) * g * t^2
d = (1/2) * 32.2 ft/s^2 * (0.4048 s)^2
d ≈ 2.066 ft
Therefore, the ball would fall approximately 2.066 feet vertically by the time it reaches home plate.