Answer :
Using differentiation, 36 feet is the maximum height, in feet, that the object will reach.
In the given question,
An object is launched directly in the air at a speed of 32 feet per second from a platform located 20 feet above the ground.
The position of the object can be modeled using the function [tex]f(t)=-16t^2+32t+20[/tex], where t is the time in seconds and f(t) is the height, in feet, of the object.
We have to find the maximum height, in feet, that the object will reach.
The given function is [tex]f(t)=-16t^2+32t+20[/tex].
To find the maximum height we differentiate the function with respect to t
[tex]\frac{df}{dt}=-32t+32[/tex]
As we know that at the maximum point the slope of the tangent is zero.
So df/dt = 0
So -32t+32=0
Subtract 32 on both side
-32t+32-32=0-32
-32t=-32
Divide by -32 on both side
-32/-32 t = -32/-32
t=1
Therefore at t=1 the function have maximum value. So we put t=1 in the given equation.
[tex]f(1)=-16(1)^2+32\times1+20[/tex]
f(1)=(-16)×1+32×1+20
Simplifying
f(1)=-16+32+20
f(1)=36 feet.
Hence, 36 feet is the maximum height, in feet, that the object will reach.
To learn more about differentiation link is here
brainly.com/question/14496325
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