Post created by: Laurissa
In baseball, there are multiple ways to hit a home run. Home runs can be arched or can be a line-drive. In other words, a line-drive home run is hit when the most direct angle to hit a ball over the fence on the opposite side of the field is achieved. In this example, we will use the San Francisco Giants' Kevin Pillar's 14th home run to analyze the exit velocity of the ball and time it takes the ball to travel out of the park. First, to calculate this, we have to create some assumptions. On average, a major league swing is 70 mph (31.29 m/s), the mass of a ball is 149 g and the average mass of a major league bat is 32 oz (907.19 g). In the example, the velocity of the pitch is 88 mph (39.34 m/s) in the opposite direction. Assuming this is an elastic collision and momentum is conserved, the exit velocity of the ball in the x-direction is 81.99 m/s.
Using the optimal angle of a line-drive home run which is between 15 - 20 degrees, The total exit velocity is between 84.89 - 87.26 m/s (189.89 - 195.20 mph). After doing research, the fastest exit velocity recorded of a home run is 121.7 mph with a 17-degree launch angle hit by Giancarlo Stanton on August 9, 2018.
The initial assumptions taken at the beginning of the problem account for this large gap in theoretical versus the actual exit velocity. First, the ball does not travel in a straight line as it travels towards the plate. There is spin on the ball that causes a two-dimensional collision and a variation in velocity. Another force that is not accounted for in this calculation is air resistance. These two ignored variables account for much of the difference between the theoretical and actual exit velocities. https://www.mlb.com/video/kevin-pillar-homers-14-on-a-line-drive-to-left-field?t=most-popular http://m.mlb.com/statcast/leaderboard#avg-pitch-velo,r,2018
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