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:
[tex]1\leq t\geq \sqrt{2}[/tex]
Step-by-step explanation:
Given:
A stuntman jumping off a 20-m-high building is modeled by the equation
[tex]h =20-5t^{2}[/tex]-----------(1)
A high-speed camera is ready to making film between 15 m and 10 m above the ground
when the stuntman is 15m above the ground.
height [tex]h = 15m[/tex]
Put height value in equation 1
[tex]15 =20-5t^{2}[/tex]
[tex]5t^{2} =20-15[/tex]
[tex]5t^{2} =5[/tex]
[tex]t^{2} =1[/tex]
[tex]t =\pm1[/tex]
We know that the time is always positive, therefore [tex]t=1[/tex]
when the stuntman is 10m above the ground.
height [tex]h = 10m[/tex]
Put height value in equation 1
[tex]10 =20-5t^{2}[/tex]
[tex]5t^{2} =20-10[/tex]
[tex]5t^{2} =10[/tex]
[tex]t^{2} =\frac{10}{5}[/tex]
[tex]t^{2} =2[/tex]
[tex]t=\pm\sqrt{2}[/tex]
[tex]t=\sqrt{2}[/tex]
Therefore ,time interval of camera film him is [tex]1\leq t\geq \sqrt{2}[/tex]
