Answer:
A)
At t = 3/2
B)
36 feet
Step-by-step explanation:
The soccer ball is modeled by the function:
[tex]h(t)=-16t^2+48t[/tex]
Where h(t) represents the height in feet of the ball over time t.
Note that this is a quadratic equation.
Part A)
Since this is a quadratic, the maximum height will be reached when the ball reaches its vertex.
So, we will find the x-coordinate of the vertex.
The vertex is given by:
[tex]\displaystyle \Big (-\frac{b}{2a} , h( -\frac{b}{2a} ) \Big)[/tex]
In this case, a = -16 and b = 48. Thus, the t-coordinate is:
[tex]\displaystyle t=-\frac{48}{2(-16)}=-\frac{48}{-32}=\frac{3}{2}[/tex]
The ball will reach is maximum height at t = 3/2.
Part B)
To find the maximum height, we can simply substitute the value back into the function and evaluated. Therefore:
[tex]\displaystyle\begin{aligned} h(\frac{3}{2})&=-16\Big(\frac{3}{2}\Big)^2+48\Big(\frac{3}{2}\Big)\\\\ &=-16\Big(\frac{9}{4}\Big)+24(3)\\ \\ &=-36+72\\\\ &= 36 \end{aligned}[/tex]
The maximum height of the ball is 36 feet.
