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.
To solve this problem, we can use the kinematic equation:
d = vit + 0.5at^2
Where:
d = distance travelled (18.3 m)
vi = initial velocity (unknown)
a = acceleration due to gravity (-9.81 m/s^2, since the ball is falling vertically)
t = time taken (unknown)
First, let's calculate the time it takes for the pitch to reach home plate:
d = 0.5at^2
18.3 = 0.5 * (-9.81) * t^2
Solving for t, we get:
t^2 = (18.3 * 2) / 9.81
t^2 = 3.73
t ≈ √3.73
t ≈ 1.93 seconds
Now that we have the time, we can use it to calculate the initial velocity:
d = vit
2.65 = vi * 1.93
Solving for vi, we get:
vi = 2.65 / 1.93
vi ≈ 1.37 m/s
So, Nolan Ryan's pitch was traveling at approximately 1.37 m/s.
To solve this problem, we can use the equations of motion. Since the ball is thrown horizontally, we only need to consider the vertical motion.
We know that the initial vertical position, y0, is 0 meters (since the ball starts at the same height as the home plate). The final vertical position, y, is -0.809 meters (since the ball falls by 0.809 meters when it reaches home plate). The initial velocity, v0, is 0 m/s (since the ball is thrown horizontally and has no vertical component initially). The acceleration, a, is -9.81 m/s^2 (since the ball accelerates downward due to gravity).
Using the equation of motion y = y0 + v0t + (1/2)at^2, we can solve for t (time of flight):
-0.809 = 0 + 0*t + (1/2)(-9.81)t^2
Rearranging the equation, we have:
4.905t^2 = -0.809
Dividing both sides of the equation by 4.905:
t^2 = -0.809 / 4.905
Taking the square root of both sides:
t ≈ √(-0.809 / 4.905)
Note that we are only considering the positive value of t since time cannot be negative. So, we have:
t ≈ 0.406 seconds
Now that we know the time of flight, we can find the horizontal velocity using the equation v = d / t, where d is the horizontal distance traveled by the ball (18.3 m in this case):
v = 18.3 m / 0.406 s
Simplifying, we find:
v ≈ 45.078 m/s
Therefore, the speed of Nolan Ryan's pitch was approximately 45.078 m/s.