Answer: [tex]1\leq t\leq \sqrt{2}[/tex]
Step-by-step explanation:
1. You know that the high-speed camera is ready to film him between 15 meters and 10 meters above the ground.
2. The problem gives you the following equation:
[tex]h=20-5t^{2}[/tex]
3. Therefore, you must calculate the time when [tex]h=15[/tex] and when [tex]h=10[/tex], as following:
[tex]h=15\\15=20-5t^{2}\\5t^{2}=5\\t=\sqrt{1}\\t=1[/tex]
[tex]h=10\\10=20-5t^{2}\\5t^{2}=10\\t=\sqrt{2}[/tex]
4. Therefore, the answer is:
[tex]1\leq t\leq \sqrt{2}[/tex]
