Respuesta :
Answer:
If you want to round to the nearest hundredths, the answer is 1.73 seconds.
Step-by-step explanation:
So we want to solve h(t)=5 for t because this will give us the time,t, that the diver was 5 m above the water.
[tex]-4.9t^2+1.5t+17=5[/tex]
My goal here in solving this equation is to get it into [tex]at^2+bt+c=0[/tex] so I can use the quadratic formula to solve it.
The quadratic formula is [tex]t=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex].
So let's begin that process here:
[tex]-4.9t^2+1.5t+17=5[/tex]
Subtract 5 on both sides:
[tex]-4.9t^2+1.5t+12=0[/tex]
So let's compare the following equations:
[tex]-4.9t^2+1.5t+12=0[/tex]
[tex]at^2+bt+c=0[/tex].
[tex]a=-4.9[/tex]
[tex]b=1.5[/tex]
[tex]c=12[/tex]
Now we are ready to insert in the quadratic formula:
[tex]t=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]t=\frac{-1.5 \pm \sqrt{(1.5)^2-4(-4.9)(12)}}{2(-4.9)}[/tex]
I know this can look daunting when putting into a calculator.
But this is the process I used on those little calculators back in the day:
Put the thing inside the square root into your calculator first. I'm talking about the [tex](1.5)^2-4(-4.9)(12)[/tex].
This gives you: 237.45
Let's show what we have so far now:
[tex]t=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]t=\frac{-1.5 \pm \sqrt{(1.5)^2-4(-4.9)(12)}}{2(-4.9)}[/tex]
[tex]t=\frac{-1.5 \pm \sqrt{237.45}}{2(-4.9)}[/tex]
I'm going to put the denominator, 2(-4.9), into my calculator now.
[tex]t=\frac{-1.5 \pm \sqrt{237.45}}{-9.8}[/tex]
So this gives us two numbers to compute:
[tex]t=\frac{-1.5 - \sqrt{237.45}}{-9.8} \text{ and } t=\frac{-1.5+\sqrt{237.45}}{-9.8}[/tex]
I'm actually going to type in -1.5-sqrt(237.45) into my calculator, as well as, -1.5+sqrt(237.45).
[tex]t=\frac{-16.90941271}{-9.8} \text{ and } t=\frac{13.90941271}{-9.8}[/tex]
We are going to use the positive number only for our solution.
So we have the answer is whatever that first fraction is approximately:
[tex]t=\frac{-16.90941271}{-9.8}=1.725450277[/tex].
The answer is approximately 1.73 seconds.
