Question

the height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 - 5t², 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?
o t = √2
o 1 < t < √2
o t > √2
o t < 2

Explanation:

Step1: Set up inequalities

We know that \(10 < h<15\). Substitute \(h = 20 - 5t^{2}\) into the inequality: \(10<20 - 5t^{2}<15\).

Step2: Split the compound - inequality

Split \(10<20 - 5t^{2}<15\) into two inequalities: \(10<20 - 5t^{2}\) and \(20 - 5t^{2}<15\).

Solve \(10<20 - 5t^{2}\):

Subtract 20 from both sides: \(10-20<20 - 5t^{2}-20\), which gives \(- 10<-5t^{2}\). Divide both sides by \(-5\) and reverse the inequality sign: \(2>t^{2}\), or \(t^{2}<2\). Taking the square - root of both sides (and considering \(t\geq0\) since \(t\) represents time), we get \(t <\sqrt{2}\).

Solve \(20 - 5t^{2}<15\):

Subtract 20 from both sides: \(20 - 5t^{2}-20<15 - 20\), which gives \(-5t^{2}<-5\). Divide both sides by \(-5\) and reverse the inequality sign: \(t^{2}>1\). Taking the square - root of both sides (and considering \(t\geq0\) since \(t\) represents time), we get \(t > 1\).

Answer:

\(1 < t<\sqrt{2}\)