Question

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.

Answer 1

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, $y=52x-16x^2$. The y-coordinate we will get will be the maximum height.

Let's start.

The x-coordinate of the vertex is $\frac{-b}{2a}$.

$y=52x-16x^2$ compare to $y=ax^2+bx+c$.

We have that $a=-16,b=52,c=0$.

Let's plug into  $\frac{-b}{2a}$ with those values.

$\frac{-b}{2a}$ with $a=-16,b=52,c=0$

$\frac{-52}{2(-16)}=\frac{52}{32}=\frac{26}{16}=\frac{13}{8}$.

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

Now to find the corresponding y-coordinate.

$y=52(\frac{13}{8})-16(\frac{13}{8})^2$

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

$y=\frac{169}{4} \text{ or } 42.25$

So the maximum is 42.25 feet.

Answer 2

Answer: 42.25 feet

Step-by-step explanation:

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

$h(t) = 52t -16t^2$

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):

$x=t=\frac{-b}{2a}$

In this case:

$a=-16\\b=52$

Substituting values, we get :

$t=\frac{-52}{2(-16)}\\\\t=1.625$

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

$h(1.625) = 52(1.625) -16(1.625)^2\\\\h(1.625) =42.25\ ft$