Question

I can't figure this question out. 50 points if you can answer this for me!

Answer 1

(a) $h(t)$ gives the height at time $t$, so the flare's starting height is given by $h(0)$:

$h(0)=-5(0)^2+90(0)+1=1\,\mathrm m$

(b) There are several ways to find the maximum height of the flare. One is to complete the square and write $h(t)$ in vertex form:

$-5t^2+90t+1=-5(t^2+18t)+1=-5(t^2-18t+81-81)+1=-5(t-9)^2+406$

That is, $h(t)$ describes a parabola whose vertex is located at (9, 406); the coefficient of -5 tells us that the parabola is concave, which means the parabola "opens" downward, and the vertex is a maximum. So the maximum height is 406 m.

(C) The flare hits the ground when $h(t)=0$:

$-5t^2+90t+1=0\implies t=9\pm\sqrt{\dfrac{406}5}$

or at about $t\approx-0.011$ and $t\approx18.01$. We ignore the negative solution (negative time makes no physical sense).