Mathematics: 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

143 divided by 2 = 71.571.5 - .5 = 71.5 + .5 = 171 + 1 = 72So the answer is 71 and 72...

The height of a stuntperson jumping off a building

Ответ:

venny23

27.07.2020

Mathematics

Time Interval : 1 sec to 1.414 sec

Step-by-step explanation:

The height of the stuntperson jumping off a $20\text{ }m$ is modelled as $h\text{ }=\text{ }20\text{ }-\text{ }5t^{2}$ . Here is the time in seconds.

We have to record the person when he is between the heights $15\text{ }m\text{ and }10\text{ }m$ from the ground.

When $h\text{ }=\text{ }15$ ,

$15\text{ }=20\text{ }-\text{ }5t^{2}\\5t^{2}\text{ }=\text{ }5\\t^{2}\text{ }=\text{ }1\\t\text{ }=\text{ }1\text{ }sec$

When $h\text{ }=\text{ }10$ ,

$10\text{ }=20\text{ }-\text{ }5t^{2}\\5t^{2}\text{ }=\text{ }10\\t^{2}\text{ }=\text{ }2\\t\text{ }=\text{ }\sqrt{2}\text{ }=\text{ }1.414\text{ }sec$

The stuntperson is at height $15\text{ }m$ at $t\text{ }=\text{ }1\text{ }sec$ , and is at height $10\text{ }m$ at $t\text{ }=\text{ }1.414\text{ }sec$ .

∴ The camera should film the person for the time interval $\textbf{1}\text{ }sec\text{ to }\textbf{1.414}\text{ }sec$ .

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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

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