Question

From home plate to the center field wall at a ball park is 130 meters. When a batter hits a long drive the ball leaves his bat 1 meter off the ground with a velocity of 40 meters per second at 28 degrees above the horizontal. The center field wall is 2.6 meters high. Does he hit a home run?

Answer 1

Answer on Question 82036, Physics, Astronomy, Astrophysics

Question:

From home plate to the center field wall at a ball park is 130 meters. When a batter hits a long drive the ball leaves his bat 1 meter off the ground with a velocity of 40 meters per second at 28 degrees above the horizontal. The center field wall is 2.6 meters high. Does he hit a home run?

Solution:

Let's, consider the motion of the ball in two dimensions:

$$x = v _ {0 x} t = v _ {0} \cos \theta t,$$ $$y = y _ {0} + v _ {0 y} t - \frac {1}{2} g t ^ {2} = y _ {0} + v _ {0} \sin \theta t - \frac {1}{2} g t ^ {2},$$

here, $x = 130 \, \text{m}$ is the horizontal distance traveled by the ball to the center field wall, $v_0 = 40 \, \text{m/s}$ is the initial velocity of the ball, $\theta = 28{}^\circ$ is the launch angle, $y$ is the height of the ball, $y_0 = 1.0 \, \text{m}$ is the initial height of the ball above the ground, $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity and $t$ is the time.

Let's find the height of the ball as a function of horizontal distance by eliminating the time. Let's express the time from the equation (1) and substitute it into the equation (2):

$$t = \frac {x}{v _ {0} \cos \theta},$$ $$y = y _ {0} + v _ {0} \sin \theta \cdot \frac {x}{v _ {0} \cos \theta} - \frac {1}{2} g \left(\frac {x}{v _ {0} \cos \theta}\right) ^ {2},$$ $$\begin{array}{l} y = y _ {0} + x \tan \theta - \frac {1}{2} g \left(\frac {x}{V _ {0} \cos \theta}\right) ^ {2} = \\ = 1. 0 m + 1 3 0 m \cdot \tan 2 8 {}^ {\circ} - \frac {1}{2} \cdot 9. 8 \frac {m}{s ^ {2}} \cdot \left(\frac {1 3 0 m}{4 0 \frac {m}{s} \cdot \cos 2 8 {}^ {\circ}}\right) ^ {2} = \\ = 1. 0 m + 2. 7 3 m = 3. 7 3 m. \\ \end{array}$$

Since, the height of the ball is greater than the height of the center field wall, the ball will clear this wall and the batter will hit a home run.

Answer:

$y = 3.73\,m$, the batter will hit the home run.

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