The maximum height in feet, after the ball was thrown is 14 feet and this can be determined by differentiating the given function.
Given :
\(h(t) = 14+32t-16t^2\)
The distance of the ball above the ground at any time 't' is given by the function:
\(h(t) = 14+32t-16t^2\) ---- (1)
Differentiate the above equation to determine the maximum height of the ball after throwing.
\(\dfrac{d(h(t))}{dt} =\dfrac{d}{dt}(14+32t-16t^2)\)
\(\dfrac{d(h(t))}{dt} = 0 +32-32t\)
Now, for maximum height, \(\dfrac{(h(t))}{dt} = 0\).
32 - 32t = 0
t = 1 second
Now, put the value of 't' in equation (1).
\(h(1) = 14+32\times1-32\times(1)^2\)+
h = 14 feet.
The maximum height in feet, after the ball was thrown is 14 feet.
For more information, refer to the link given below:
https://brainly.com/question/14496325
