The fastest recorded pitch in the MLB was thrown by Nolan Ryan in 1974. If this pitch were thrown horizontally, the ball would falll .809m (2.65 ft) by the time it reached home plate, 18.3 m away (60 ft).
The acceleration of gravity is 9.81 m/s^2. How fast was Ryan's pitch? Answer in units of m/s.
45.07
To find the velocity of Nolan Ryan's pitch, we can use the equation of motion that relates distance, time, initial velocity, and acceleration:
distance = (initial velocity × time) + (0.5 × acceleration × time^2)
In this case, the distance is 18.3 m, the acceleration is the acceleration due to gravity (9.81 m/s^2), and the distance fallen (2.65 ft) can be considered as the initial displacement of the pitch, set to zero velocity.
Thus, our equation becomes:
0.809 m = (0 × t) + (0.5 × 9.81 m/s^2 × t^2)
Simplifying the equation, we get:
4.905 t^2 = 0.809
Divide both sides of the equation by 4.905:
t^2 = 0.165311
To isolate t, take the square root of both sides:
t = sqrt(0.165311)
t ≈ 0.4067 s
Now that we have the time it took for the pitch to reach home plate, we can find the velocity using the equation:
velocity = distance / time
velocity = 18.3 m / 0.4067 s
velocity ≈ 45.02 m/s
Therefore, the speed of Nolan Ryan's pitch, if thrown horizontally, was approximately 45.02 m/s.