Part A:
Given that the height of a batted ball is modeled by the function [tex]h=-0.01x^2+1.22x+3[/tex], where x is the horizontal distance in feet from the point of impact with the bat, and h is the height of the ball in feet.
At maximum height,
[tex]\frac{dh}{dx} =0[/tex]
[tex]\frac{dh}{dt} =0 \\ \\ \Rightarrow-0.02x+1.22=0 \\ \\ \Rightarrow0.02x=1.22 \\ \\ \Rightarrow x= \frac{1.22}{0.02} =61[/tex]
Thus, the maximum height occurs when x = 61 feet and the maximum height is given by
[tex]h=-0.01(61)^2+1.22(61)+3 \\ \\ =-0.01(3,721)+74.42+3=-37.21+77.42 \\ \\ =40.21[/tex]
Therefore, the maximum height is 40.21 feet.
Part B:
From part A, we can see that the maximum height occured when x = 61 feet.
Therefore, the distance from the batter the ball will be at its maximum height is 61 feet.
