Answer:
2.83 s
Step-by-step explanation:
We are given that
[tex]h(t)=-16t^2+10t+100[/tex]
When a rock reached the ground then h(t)=0
[tex]-16t^2+10t+100=0[/tex]
[tex]16t^2-10t-100=0[/tex]
Using quadratic formula
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]t=\frac{10\pm\sqrt{(-10)^2-4\times 16\times (-100)}}{2(16)}[/tex]
[tex]t=\frac{10\pm\sqrt{100+6400}}{32}[/tex]
[tex]t=\frac{10+\sqrt{6500}}{32}=2.83[/tex]
[tex]t=\frac{10-\sqrt{6500}}{32}=-2.2[/tex]
Time cannot be negative .Therefore, negative value of t can not be possible.
Hence, the rock takes 2.83 sec to reach the ground.
