I'm sitting on my warm, snuggly couch watching the Cleveland Indians and the Chicago Cubs in Game 2 of the World Series. The players, coaches, and fans are shivering on my television screen. Today, we are going to discuss the physics of home runs; unfortunately, the cold weather makes it unlikely anyone will hit a dinger in tonight's game. This phenomenon is rooted in fluid dynamics. Cold air has a higher density which results in increased drag on the ball. However, that discussion is outside the scope of this article. We are going to investigate how launch angle impacts whether you flyout or jog around the bases. Recall that the trajectory of a falling object is parabolic; meaning, it has an x and a ycomponent. Knowing that distance travelled by a falling object is related to the initial velocity (v), we can mathematically express each component of the velocity utilizing concepts from trigonometry: We get the following expressions after plugging the expressions above into the formulas for a zero launch angle: Now we can really start digging into our problem. Let's make the following assumptions:
You may notice we don't have a time component, which is a problem since we have 3 unknowns and two equations. However, we can overcome this by reframing our question: How far has the ball travelled when it reaches a height of 3 m? We ask this question because we want to find the ideal launch angle to hit a home run. So the mechanics look like this: Therefore, under the assumed conditions a launch angle of 45 degrees would result in a home run! Let's look at a range of angles to find the "sweet spot:" It appears when the ball is traveling at 35 m/s the ideal launch angle is between 45 and 50 degrees. The next question is: How would this change with varying initial velocities? As expected, you have greater flexibility of launch angle with increasing velocity. So if you want to crush the ball, you better start pumping iron!
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AuthorEnjoys reading, listening to TedTalks, and discussing new concepts with others. Archives
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