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Mathematics 20 Online
OpenStudy (cutiecomittee123):

A stranded soldier shoot a signal flare into the air to attract the attention of a nearby plan. The flare has an initial velocity of 1500 feet per second. Its height is defined by the quadratic function below. Assume that the flare is fired from ground level.

OpenStudy (cutiecomittee123):

I forgot to add the questions; What is the maximum height that the flare reaches? Show your work When will the flare reach that height? Show your work At what time does the flare hit the ground again? Show your work and if the plan is flying at a height of 30,000 feet, at a speed of 880 per second, and it is 50,000 feet from the flare, will the flare hit it? If so, tell when this will happen. If not when the flare reaches the planes altitude.

OpenStudy (phi):

**the quadratic function below** without this info, you can't answer the question.

OpenStudy (cutiecomittee123):

the quadratic function below is h=vit-16r^2

OpenStudy (phi):

do you mean \[ h = v_i t - 16t^2 \] no r, right?

OpenStudy (cutiecomittee123):

Yes!

OpenStudy (cutiecomittee123):

Sorry this is my first post here and im confused

OpenStudy (phi):

\( v_i\) is short for velocity_initial, or 1500 ft/sec

OpenStudy (cutiecomittee123):

okay

OpenStudy (cutiecomittee123):

so to find the maximum height... can i just graph it?

OpenStudy (cutiecomittee123):

But how would that be graphed on a graphing calculator

OpenStudy (phi):

so your equation is \[ h = 1500 t - 16t^2\] you can write it this way \[ h = -16t^2 +1500 t \] in case this does not ring any bells, it's a parabola

OpenStudy (phi):

do you know how to find the "vertex" (peak) of a parabola? (graphing, unless with a computer, is harder than using algebra for this question)

OpenStudy (cutiecomittee123):

Okay that makes sense. and then to type it into a calculator you would just do y=-16x^2+1500x then just find the highest point, inorder to find the maximum height?

OpenStudy (phi):

yes

OpenStudy (cutiecomittee123):

I do recognize that it is a parabola by looking at the equation

OpenStudy (phi):

if you have a x^2 + bx + c the peak is at x = -b/2a in your case at t= -1500/-32

OpenStudy (cutiecomittee123):

So then the peak is 46.875

OpenStudy (cutiecomittee123):

which is the highest point?

OpenStudy (phi):

t= 46.875. that is when the flare reaches the max height (Q2) to find the height , use that number in 1500t - 16t^2

OpenStudy (cutiecomittee123):

got it so now how do i figure out when the flare hit the ground again? would that just be the second x intercept?

OpenStudy (phi):

yes. it's where y=0 which happens at t=0 (going up) and at t= whatever,when it lands but we know it's symmetric, so it's 2*t where t is the time to the peak

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