Question

PLEASE HELP!! I've stayed up for 3 hours trying to solve this one problem! I just can't figure this problem out.

Use the quadratic formula (x=-b±√b²-4ac /2) to solve the equation. Then graph the equation. Show all work:

A flare is fired straight up from the ground with an initial velocity of 100 feet per second. How long will it take for the flare to reach an altitude of 150 feet. HINT: use formula h=-16t+vot+ho.

Answer 1

please help mane1

Answer 2

anyone

Answer 3

So far, I have done this:

h=-16t^2=vot+ho
150=-16t^2+100t+0
A:-16 B: 100 C: 150

To find the axis of symmetry the formula is x=-b/2a
x=-100/-32
x=3.125

h=-16(3.125)^2 + 100(3.125) - 150
-16(9.77) + 312.5 - 150
-156.32 + 312.5 - 150
h=6.18

However that is VERY improbable....and so I believe I went wrong somewhere however I can't pinpoint exactly where I went wrong....any suggestions?

Answer 4

\[h(t) = -16t^2+v_0 t + h_0\]\[h(t) = -16t^2+100t + 0\]\[150 = -16t^2+100t\]Now we need to solve for the value of \(t\) that makes that happen. There's nothing that says it is necessarily the vertex, so finding the vertex/axis of symmetry isn't useful. Instead, just use the quadratic solution formula, or factor it or complete the square or whatever your favorite method is.

\[-16t^2+100t-150 = 0\]\[t = \frac{-100\pm\sqrt{100^2-4(-16)(-150)}}{2*(-16)} = \]You're only interested in the solution that gives you a positive value for \(t\).

As far as going astray, the main problem was thinking the vertex had something to do with the problem here. But a minor problem was when you wrote \[h=-16(3.125)^2 + 100(3.125) - 150\]That -150 has no business being here — it only comes about because you are trying to find the value of \(t\) such that \(h(t) = 150\). By writing \[h = -16(3.125)^2 + 100(3.125) - 150\]you have put the height in the formula in two different places. You might not have been so easily led astray had you used \(h(t)\) instead of \(h\), making it clearer that you are talking about a value of a function at a specific point.