The maximum horizontal distance traveled by a ball that leaves a bat and returns to the ground level is 145.8 meters.
The maximum horizontal distance ([tex]x_{max}[/tex]) of the ball when it returns to the ground level can be calculated with the following equation:
[tex] x_{max} = \frac{v_{0}^{2}sin(2\theta)}{g} [/tex]
Where:
[tex]v_{0}[/tex]: is the initial speed = 38 m/s
θ: is the angle = 41° above horizontal
g: is the acceleration due to gravity = 9.81 m/s²
Hence, the maximum horizontal distance is:
[tex]x_{max} = \frac{v_{0}^{2}sin(2\theta)}{g} = \frac{(38 m/s)^{2}sin(2*41)}{9.81 m/s^{2}} = 145.8 m[/tex]
Therefore, the horizontal distance when the ball returns to ground level is 145.8 meters.
You can learn more about the maximum horizontal distance equation here: https://brainly.com/question/14701768?referrer=searchResults
I hope it helps you!
