Question

a toy rocket is fired into the air from the top of a barn. its height (h) above the ground in yards after t seconds is given by the function $h(t) = -5t^2 + 10t + 20$.

1. what was the maximum height of the rocket?
2. how long was the rocket in the air before hitting the ground?
3. at what time(s) will the rocket be at a height of 22 yd?

Explanation:

Step1: Find time of max height

For quadratic $h(t)=at^2+bt+c$, vertex time is $t=-\frac{b}{2a}$.
Here $a=-5$, $b=10$, so $t=-\frac{10}{2(-5)} = 1$ second.

Step2: Calculate max height

Substitute $t=1$ into $h(t)$.
$h(1)=-5(1)^2+10(1)+20 = -5+10+20 = 25$

Step3: Find time to hit ground

Set $h(t)=0$, solve $-5t^2+10t+20=0$.
First divide by $-5$: $t^2-2t-4=0$. Use quadratic formula $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=1$, $b=-2$, $c=-4$.
$t=\frac{2\pm\sqrt{(-2)^2-4(1)(-4)}}{2(1)}=\frac{2\pm\sqrt{4+16}}{2}=\frac{2\pm\sqrt{20}}{2}=\frac{2\pm2\sqrt{5}}{2}=1\pm\sqrt{5}$.
Discard negative time: $t=1+\sqrt{5}\approx3.24$ seconds.

Step4: Find time for height 22 yd

Set $h(t)=22$, solve $-5t^2+10t+20=22$.
Rearrange: $-5t^2+10t-2=0$, divide by $-1$: $5t^2-10t+2=0$.
Quadratic formula: $t=\frac{10\pm\sqrt{(-10)^2-4(5)(2)}}{2(5)}=\frac{10\pm\sqrt{100-40}}{10}=\frac{10\pm\sqrt{60}}{10}=\frac{10\pm2\sqrt{15}}{10}=1\pm\frac{\sqrt{15}}{5}$.
Simplify: $t=1+\frac{\sqrt{15}}{5}\approx1.77$ s, $t=1-\frac{\sqrt{15}}{5}\approx0.23$ s.

Answer:

1. 25 yards
2. $1+\sqrt{5}$ seconds (or approximately 3.24 seconds)
3. $1-\frac{\sqrt{15}}{5}$ seconds and $1+\frac{\sqrt{15}}{5}$ seconds (or approximately 0.23 seconds and 1.77 seconds)