The maximum height of the baseball is about 90 feet
<h3>Further explanation</h3>
Acceleration is rate of change of velocity.


<em>a = acceleration ( m/s² )</em>
<em>v = final velocity ( m/s )</em>
<em>u = initial velocity ( m/s )</em>
<em>t = time taken ( s )</em>
<em>d = distance ( m )</em>
Let us now tackle the problem!
<u>Given:</u>
The height s of the ball in feet is given by the equation:

To find the velocity function, the above equation will be derived as follows:



At the maximum height, the speed is 0 m/s , then:









<h3>Learn more</h3>
<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle