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

Answer 1

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 $t$ 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 $\textbf{1}\text{ }sec\text{ to }\textbf{1.414}\text{ }sec$.

Answer 2

Answer:

A

Step-by-step explanation: