Math algebra 2 What is the max height and seconds.

Question

A A diver leaps into the sea from the cliff edge. His height from the water surface as a function of time is modeled by the following parabola:h(t) = -5t^2+5t+12. Determine how long in seconds the driver was airborne before striking the water. Also determine the maximum height reached during his jump and meters

Answer 1

The diver will strike the water when h(t) = 0. Use the quadratic equation to solve for t and take the positive value.

t = (-5±√(25-4(-5)(12))/-(10) = -1.27, 2.13 pick 2.13 seconds

For a parabola, the vertex is t=-b/2a = -5/(2(-10)) = 1/2

So, h(1/2) = -5(1/2)² + 5(1/2) + 12 = 13.25 meters

Answer 2

The time it takes for the diver to hit the water is obtained by solving for t when h = 0 (elevation of the water). So you're looking at:

-5t^2 + 5t + 12 = 0

Use the quadratic equation to solve this. If ax^2 + bx + c = 0, then:

x = (-b +- sqrt(b^2 - 4ac))/2a

When you apply it to this problem, a = -5, b = 5 and c = 12. Be sure to choose the positive value for t!

To get the max height you should first graph the function--this will give you an approximate answer (the value of the parabola at its peak). But to get the exact answer, you're going to want to "complete the square."

First rewrite the equation as t^2 - t = (12 - hmax) / 5. This is simple rearrangement. hmax is the maximum height of the dive.

We want to rewrite the lefthand side as (t - a)^2 + b. Then we will have "completed the square." Why? Because once we've done that, we can solve the equation for hmax. When you do so, you'll find that the multiplier of (t - a)^2 is negative, i.e., hmax = constants - d * (t - a)^2. Because (t - a)^2 can never be negative, we make hmax as large as possible by choosing t = a so this term drops out. The result is the max height.