Respuesta :
Answer:
Time taken to reach maximum height=0.875 seconds
Maximum height = 78.25 ft
Step-by-step explanation:
The function which models the height of the ball is given as:[tex]s(t)=-16t^2+28t+66[/tex]
The function s(t) reaches its maximum height at the axis of symmetry. For a quadratic equation of the form [tex]ax^2+bx+c=0[/tex], the axis of symmetry occurs at [tex]x=-\frac{b}{2a}[/tex].
In s(t), a=-16, b=28
[tex]t=-\frac{28}{2(-16)}=0.875 seconds[/tex]
The ball reaches its maximum height after 0.875 seconds
(b)Maximum Height
Given [tex]s(t)=-16t^2+28t+66[/tex]
At t=0.875
[tex]s(0.875)=-16(0.875)^2+28(0.875)+66=78.25 ft[/tex]
Maximum height = 78.25 ft
Answer:
time: 0.875 seconds
maximum height: 18.25 feet
Step-by-step explanation:
The function that gives the height of the ball in feet after t seconds is:
s(t) = −16t^2 + 28t + 6
The inicial velocity is 28 ft/sec, and the inicial position is 6 feet.
So, to find the time when the ball reaches the maximum heigth, we need to find the vertix of the equation −16t^2 + 28t + 6.
We can use the following formula to find it:
t_vertix = -b/2a
where a and b are coefficients of the quadratic equation (in our case, a = -16 and b = 28).
So, we have that:
t_vertix = -28/(-32) = 0.875 seconds
To calculate the maximum height we just need to use this time in the equation of position:
s(0.875) = −16*(0.875)^2 + 28*0.875 + 6 =  18.25 feet
