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A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h(t) = 52t - 16t^2 . What is the maximum height that the ball will reach?
Do not round your answer.

Respuesta :

luisejr77

Answer: 42.25 feet

Step-by-step explanation:

We know that after "t" seconds, its height "h" in feet is given by this function:

[tex]h(t) = 52t -16t^2[/tex]

The maximum height is the y-coordinate of the vertex of the parabola. Then, we can use the following formula to find the corresponding value of "t" (which is the x-coordinate of the vertex):

[tex]x=t=\frac{-b}{2a}[/tex]

In this case:

[tex]a=-16\\b=52[/tex]

Substituting values, we get :

[tex]t=\frac{-52}{2(-16)}\\\\t=1.625[/tex]

Substituting this value into the function to find the maximum height the ball will reach, we get:

[tex]h(1.625) = 52(1.625) -16(1.625)^2\\\\h(1.625) =42.25\ ft[/tex]

Answer:

42.25 feet

Step-by-step explanation:

The maximum of a quadratic can be found by finding the vertex of the parabola that the quadratic creates visually on a graph.

So first step to find the maximum height is to find the x-coordinate of the vertex.

After you find the x-coordinate of the vertex, you will want to find the y that corresponds by using the given equation, [tex]y=52x-16x^2[/tex]. The y-coordinate we will get will be the maximum height.

Let's start.

The x-coordinate of the vertex is [tex]\frac{-b}{2a}[/tex].

[tex]y=52x-16x^2[/tex] compare to [tex]y=ax^2+bx+c[/tex].

We have that [tex]a=-16,b=52,c=0[/tex].

Let's plug into  [tex]\frac{-b}{2a}[/tex] with those values.

[tex]\frac{-b}{2a}[/tex] with [tex]a=-16,b=52,c=0[/tex]

[tex]\frac{-52}{2(-16)}=\frac{52}{32}=\frac{26}{16}=\frac{13}{8}[/tex].

The vertex's x-coordinate is 13/8.

Now to find the corresponding y-coordinate.

[tex]y=52(\frac{13}{8})-16(\frac{13}{8})^2[/tex]

I'm going to just put this in the calculator:

[tex]y=\frac{169}{4} \text{ or } 42.25[/tex]

So the maximum is 42.25 feet.

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