A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?
Answer: Time interval of camera film him is \(1\leq t\leq \sqrt{2}\)
Step-by-step explanation:
Given: A stuntman jumping off a 20-m-high building is modeled by the equation \(h=20-5t^2\), where t is the time in seconds..........(1)
Height of camera to film stunt man is between 15 m and 10 m.
To find interval of time the camera film him, substitute h=15m in (1),we get
\(15=20-5t^2\\\Rightarrow-5=-5t^2\\\Rightarrow\ t^2=1\\\Rightarrow\ t=1\text}[time is always positive]{\)
Now, substitute h=10m in (1),we get
\(10=20-5t^2\\\Rightarrow-10=-5t^2\\\Rightarrow\ t^2=2\\\Rightarrow\ t=\sqrt{2}\text}[time is always positive]{\)
\(\Rightarrow1\leq t\leq \sqrt{2}\)
Therefore ,time interval of camera film him is \(1\leq t\leq \sqrt{2}\)