# Aviation A helicopter hove 40 ft above the ground. Then the helicopter climbs at a rate of 21 ft/s. Write a rule that represents the helicopter's height h above the ground as a function of time t. What is the helicopter's height after 45 s?

Nov 24, 2016

${h}_{t} = 40 + 21 t$

${h}_{45} = 40 + \left(21 \times 45\right) \text{ "=" } 985 f e e t$

#### Explanation:

Let height increase be positive value
Let height decrease be negative value
Let time in seconds be ..............................$t$
Let height in feet at any time $t$ be ............${h}_{t}$

color(brown)("Initial condition")" "-> height =40 feet

Given: height increase rate is 21 feet per second $\to 21 \frac{f t}{s}$
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$\textcolor{b l u e}{\text{Building the equation}}$

After 1 second the height increases by 21 feet
After 2 seconds the height has increased by $21 + 21$ feet
After 3 seconds the height has increased by $21 + 21 + 21$ feet

So after $t$ seconds the height has $\underline{\text{changed}}$ by $21 \times t$

So we have:

height at t seconds = initial height + change in height upwards

$\text{ } \textcolor{b l u e}{{h}_{t} = 40 + 21 t}$
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$\textcolor{b l u e}{\text{Determine height after 45 seconds}}$

So for 45 seconds we have:

Note that height at 45 seconds is written as ${h}_{45}$

" "color(blue)(h_(45)=40+(21xx45)" "=" "985 feet)