Question

Identify a major league ballpark in which the distance from home plate to the center field fence and the height of the center field fence require that a ball hit 2 ft above the ground will necessitate an angle of elevation less than 0.86 to just clear the center field fence.

Please also tell me the website you access this information from.


(I need a major league ballpark with these specifications, not the formula, please!)

Answer 1

A ball hit from the home plate at the US Cellular field known formerly as Comiskey Park with a velocity of 114.72 ft/s at an angle of 0.85 radians, will just clear the center field fence which is 8 feet high and 400 ft distance from the home plate

Explanation:

Here we have the US Cellular Field

Distance of center field from the home plate = 400 ft

Height of the center field fence = 8 ft

Angle of elevation, θ = 0.85 < 0.86 r = 49.3°

The formula for projectile motion is as follows

y = y₀ + v₀ × sin(θ) × t -1/2·g·t²

x = x₀ + v₀ × cos(θ) × t

Where:

x₀ = 0

y₀ = 2 ft

x = 400 ft

y = 8 ft

g = 32.2 ft/s²

Plugging the values, we have;

8 = 2 + v₀ × sin(0.85) ×t - 1/2×32.2×t² = 2 + v₀×t × sin(0.85) - 1/2×32.2×t² ...(1)

400 = 0 + v₀ × cos(0.85) × t

Solving gives v₀×t = 606.1

Substituting the value of v₀×t in equation (1) gives t= ±5.283 s

We take the +ve value for t, hence

v₀ = 606.1/t = 606.1/5.283= 114.72 ft/s

Therefore, a ball hit from the home plate at the US Cellular field with a velocity of 114.72 ft/s at an angle of 0.85 radians, will just clear the center field fence which is 8 feet high and 400 ft distance from the home plate.

Accessed from the article on "MLB Ballparks: 9 Shortest Porches in Major League Baseball" by Sean Zerillo