Question

A cliff diver jumps from a ledge 48 feet above the ocean with an initial upward velocity of 8 feet per second. The distance the cliff diver is from the ocean surface in terms of elapsed time can be modeled by the formula d = –16t2 + 8t + 48.
When will the diver be 24 feet from the water?


How long will it take until the diver enters the water?


Explain how to find the roots and give the values.


What is the Vertex?


Explain how to find the Axis of Symmetry using your answer from the vertex.

Answer 1

Answer:

$1.5\ \text{s}$

$2\ \text{s}$

$0.25$

$(0.25,49)$

Step-by-step explanation:

The equation is

$d=-16t^2+8t+48$

When the diver be 24 feet from the water $d=24\ \text{ft}$

$24=-16t^2+8t+48\\\Rightarrow -16t^2+8t+24=0\\\Rightarrow t=\frac{-8\pm \sqrt{8^2-4\left(-16\right)\times 24}}{2\left(-16\right)}\\\Rightarrow t=-1,1.5$

So, at $t=1.5\ \text{s}$ the diver will be 24 feet away from the water.

When the diver will enter water $d=0$

$0=-16t^2+8t+48\\\Rightarrow t=\frac{-8\pm \sqrt{8^2-4\left(-16\right)\times 48}}{2\left(-16\right)}\\\Rightarrow t=-1.5,2$

So, the diver will enter the water at $t=2\ \text{s}$

The roots of the equation gives the value as in those points the distance to the water is zero.

Vertex of the curve

$t=-\dfrac{b}{2a}\\\Rightarrow t=-\dfrac{8}{2\times-16}\\\Rightarrow t=0.25$

The vertex of the curve is at $0.25$

The vertex of the curve is the highest point and the axis of symmetry passes through the vertex

$d=-16t^2+8t+48=-16\times 0.25^2+8\times 0.25+48\\\Rightarrow d=49$

So, the axis of symmetry passes through the curve at $(0.25,49)$.