A pebble is tossed into the air from the top of a cliff. The height, in feet, of the pebble over time is modeled by the equation y = -16x^2 + 32x + 80. What is the maximum height, in feet, reached by the pebble?

The maxium height is _____ feet.

A pebble is tossed into the air from the top of a cliff The height in feet of the pebble over time is modeled by the equation y 16x2 32x 80 What is the maximum class=

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Answer:

96

Step-by-step explanation:

The maximum height reached by the pebble modeled by the quadratic function,

[tex]h(t)=-16t^2+32t+80[/tex], can be found by finding the vertex.

Let's first find the t-coordinate of the vertex . The max height will correspond to this value of t which means we have to find the h(t)-coordinate.

When comparing [tex]h[/tex] to [tex]at^2+bt+c[/tex], we see that:

[tex]a=-16[/tex]

[tex]b=32[/tex]

[tex]c=80[/tex]

We need to evaluate the following to find the t-coordinate of the vertex:

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

[tex]t=\frac{-32}{2(-16)}[/tex]

[tex]t=\frac{-32}{-32}[/tex]

[tex]t=1[/tex]

So now to find the correspond h(t)-coordinate, we will need to replace t in [tex]-16t^2+32t+80[/tex] with 1:

[tex]-16(1)^2+32(1)+80[/tex]

[tex]-16(1)+32+80[/tex]

[tex]-16+32+80[/tex]

[tex]16+80[/tex]

[tex]96[/tex]

The maximum height reached by the pebble is 96 feet.

The given equation of motion;

y = -16x² + 32x + 80

At maximum height the final velocity of the pebble is zero.

The velocity of the pebble will be obtained by taking the derivative of the given equation of motion as shown below;

[tex]v =y'= \frac{dy}{dx}[/tex]

y' = -32x + 32

At maximum height, y' = 0

0 = -32x + 32

32x = 32

x = 1

The maximum height reached by the pebble is calculated as;

y(1) = -16(1)² + 32(1) + 80

y(1) = -16 + 32 + 80

y(1) = 96 feet

Thus, the maximum height reached by the pebble is 96 feet.

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