Question

the height of a stuntperson jumping off a building that is 20 m high is modeled by the equation $h = 20 - 5t^2$, where $t$ is the time in seconds. a high - speed camera is ready to film the person between 15 m and 10 m above the ground. for which interval of time should the camera film the person?\
$t = \sqrt{2}$\
$1 < t < \sqrt{2}$\
$t > \sqrt{2}$\
$t = 2$

Explanation:

Step1: Set up inequalities for height

We know \( h = 20 - 5t^2 \), and we want \( 10 \leq h \leq 15 \). So substitute \( h \) into the inequalities:
First, for the lower bound (\( h \geq 10 \)):
\( 20 - 5t^2 \geq 10 \)
Subtract 20 from both sides: \( - 5t^2 \geq 10 - 20=-10 \)
Divide both sides by - 5 (remember to reverse the inequality sign): \( t^2 \leq \frac{- 10}{-5} = 2 \)
Take square roots (since \( t\geq0 \) as it's time): \( t \leq \sqrt{2} \)

Second, for the upper bound (\( h \leq 15 \)):
\( 20 - 5t^2 \leq 15 \)
Subtract 20 from both sides: \( - 5t^2 \leq 15 - 20=-5 \)
Divide both sides by - 5 (reverse the inequality sign): \( t^2 \geq \frac{-5}{-5}=1 \)
Take square roots (since \( t\geq0 \)): \( t \geq 1 \)

Step2: Combine the inequalities

From the two steps above, we have \( 1\leq t\leq\sqrt{2} \) (or \( 1 < t < \sqrt{2} \) if we consider the open interval for the filming between the two heights, as the endpoints are the exact heights of 15m and 10m, and we want the interval between them).

Answer:

\( 1 < t < \sqrt{2} \) (assuming the option is written as \( 1 < t < \sqrt{2} \), match the option format from the problem)