A softball is tossed into the air upward from a first floor balcony. The distance of the ball 14 +32t -16t^2 above the ground at any time is given by the function, the softball above the ground (in feet) and t is the time (in seconds) what was the maximum in feet, after t was thrown? where h() is the height of

you can either find the roots of the quadratic formula using the quadratic formula two find the halfway point or you can use the formula t=-b/2a. Both of the two methods gives you the axis of symmetry (the t value for the vertex).
I will use the second method since I don't like using the quadratic formula.
t=-32/(2x-16)
t=1 second
After you find the t value for the axis of symmetry, you plug it into the formula of the graph to find the height.
h=-16+32+14
h=30 feet

ahirohit963 ahirohit963 · Answer 2 · 2021-11-11T12:55:09+01:00

The maximum height in feet, after the ball was thrown is 14 feet and this can be determined by differentiating the given function.

Given :

[tex]h(t) = 14+32t-16t^2[/tex]

The distance of the ball above the ground at any time 't' is given by the function:

[tex]h(t) = 14+32t-16t^2[/tex] ---- (1)

Differentiate the above equation to determine the maximum height of the ball after throwing.

[tex]\dfrac{d(h(t))}{dt} =\dfrac{d}{dt}(14+32t-16t^2)[/tex]

[tex]\dfrac{d(h(t))}{dt} = 0 +32-32t[/tex]

Now, for maximum height, [tex]\dfrac{(h(t))}{dt} = 0[/tex].

32 - 32t = 0

t = 1 second

Now, put the value of 't' in equation (1).

[tex]h(1) = 14+32\times1-32\times(1)^2[/tex]+

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