Answer:
36 feet.
Step-by-step explanation:
We have been given that a ball is thrown upward from ground level. Its height h, in feet, above the ground after t seconds is [tex]h(t)=-48t -16t^2[/tex]. We are asked to find the maximum height of the ball.
We can see that our given equation is a downward opening parabola, so its maximum height will be the vertex of the parabola.
To find the maximum height of the ball, we need to find y-coordinate of vertex of parabola.
Let us find x-coordinate of parabola using formula [tex]x=-\frac{b}{2a}[/tex].
[tex]x=-\frac{-48}{2(-16)}[/tex]
[tex]x=-\frac{48}{32}[/tex]
[tex]x=-\frac{3}{2}[/tex]
So, the x-coordinate of the parabola is [tex]-\frac{3}{2}[/tex]. Now, we will substitute [tex]x=-\frac{3}{2}[/tex] in our given equation to find y-coordinate of parabola.
[tex]h(t)=-48t -16t^2[/tex]
[tex]h(-\frac{3}{2})=-48(-\frac{3}{2})-16(-\frac{3}{2})^2[/tex]
[tex]h(-\frac{3}{2})=-24(-3)-16(\frac{9}{4})[/tex]
[tex]h(-\frac{3}{2})=72-4*9[/tex]
[tex]h(-\frac{3}{2})=72-36[/tex]
[tex]h(-\frac{3}{2})=36[/tex]
Therefore, the maximum height of the ball is 36 feet.
