Answer :
The height of the ball when it crosses home plate is approximately 3.4855 feet.
To find out how high the ball is when it crosses home plate, we can use the kinematic equations of projectile motion.
Given:
- Initial velocity (v₀) = 98.3 feet per second
- Launch angle (θ) = 5.2 degrees
- Initial height (y₀) = 2.7 feet
- Distance from pitcher's mound to home plate (horizontal distance, x) = 43 feet
- We are looking for the final height (y) when the ball crosses home plate.
First, we can find the time it takes for the ball to travel from the pitcher's mound to home plate using the horizontal distance and the horizontal component of velocity:
[tex]\[ x = v_0 \cdot t \cdot \cos(\theta) \][/tex]
[tex]\[ t = \frac{x}{v_0 \cdot \cos(\theta)} \][/tex]
Now, we can use this time to find the vertical position (height) of the ball using the vertical component of velocity and the time:
[tex]\[ y = y_0 + v_{0y} \cdot t - \frac{1}{2} g t^2 \][/tex]
Where:
- [tex]\( v_{0y} \)[/tex] is the vertical component of the initial velocity, which can be found using the launch angle:
[tex]\[ v_{0y} = v_0 \cdot \sin(\theta) \][/tex]
- [tex]\( g \)[/tex] is the acceleration due to gravity, which is approximately [tex]\( 32.2 \, \text{feet/second}^2 \)[/tex].
Let's plug in the given values and solve for [tex]\( y \)[/tex]:
[tex]\[ t = \frac{43}{98.3 \cdot \cos(5.2^\circ)} \][/tex]
[tex]\[ t \approx \frac{43}{98.3 \cdot 0.9962} \][/tex]
[tex]\[ t \approx \frac{43}{97.79} \][/tex]
[tex]\[ t \approx 0.4399 \, \text{seconds} \][/tex]
[tex]\[ v_{0y} = 98.3 \cdot \sin(5.2^\circ) \][/tex]
[tex]\[ v_{0y} \approx 98.3 \cdot 0.09017 \][/tex]
[tex]\[ v_{0y} \approx 8.855 \, \text{feet/second} \][/tex]
[tex]\[ y = 2.7 + (8.855 \cdot 0.4399) - \frac{1}{2} \cdot 32.2 \cdot (0.4399)^2 \][/tex]
[tex]\[ y \approx 2.7 + (8.855 \cdot 0.4399) - \frac{1}{2} \cdot 32.2 \cdot (0.4399)^2 \][/tex]
[tex]\[ y \approx 2.7 + 3.8925 - \frac{1}{2} \cdot 32.2 \cdot 0.1934 \][/tex]
[tex]\[ y \approx 2.7 + 3.8925 - 3.107 \][/tex]
[tex]\[ y \approx 3.4855 \, \text{feet} \][/tex]
Therefore, the height of the ball when it crosses home plate is approximately 3.4855 feet.
