Honors Algebra Question

Question

Robbie has a model rocket. The rocket's path is modeled by the equation y= -16x^2+80x where y + height of rocket in feet and x= time in seconds. During what time period is the rocket more than 50 feet in the air?

Answer 1

.

The vertex of the pararbola is( 2.5, 50 ) therefore the rocket never reaches a height more than 5o feet in the air.

 

It does hit the ground at .732 and and 4,268 seconds. The vertex means that at 2,5 seconds the rocket reaches its maximun height 

is 50 feet.

Answer 2

So here's what we know:

50 = -16x² + 80x (y is the beginning and end periods of the rocket hitting 50 feet)

 

If we subtract 50 from both sides, we get a quadratic:

0 = -16x² + 80x - 50

There doesn't seem to be any easy or even way to factor this out, so let's use the quadratic formula:

x = (-b ± √(b² -4ac)) / 2a

 

Now we insert the respective constants and simplify:

(-80 ± √(80² - (4 * -16 * -50))) / (2 * -16)

(-80 ± √(6400 - 3200)) / -32

(-80 ± √3200) / -32

(-80 ± 56.568) / -32

 

Now we split the ± into the addition/subtraction signs:

-80 + 56.568 = -23.432

-80 - 56.568 = -136.568

 

Now, we finally divide that denominator:

-23.432 / -32 ≈ 0.732

-136.568 / -32 ≈ 4.268

 

So the rocket should be above 50 feet between approximately 0.732 and 4.268 seconds post launch.