A cliff diver dives from 17m above the water. The diver’s height above the water, h(t) in metres after t seconds is modelled by h(t) = -4.9t2 + 1.5t + 17. Determine when the diver was 5 m above the water.Please help :(

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:  237.45Let'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.