Answer:
At t = 2 maximum height achieved by the ball is 32 feet.
Step-by-step explanation:
The height of a soccer ball can be modeled by the function h(t) = -8t² + 32t
where h(t) is the height and t is the time.
For maximum height we will find the derivative of h(t) and equate the derivative to zero.
[tex]\frac{d(h)}{dt}=\frac{d(-8t^{2}+32t)}{dt}[/tex]
[tex]\frac{d(h)}{dt}[/tex] = -16t + 32
Now we will equate derivative [tex]\frac{d(h)}{dt} = 0[/tex]
So -16t + 32 = 0
16t = 32
t = 2 seconds
At t = 2 seconds height achieved by the ball will be maximum.
Now we can calculate height h at t = 2 seconds
h(2) = -8×(2)² + 32×(2)
h(2) = -32 + 64
h(2) = 32 feet
Therefore, at t = 2 seconds soccer ball achieves the maximum height of 32 feet.
