Question

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 1

If you want to round to the nearest hundredths, the answer is 1.73 seconds.

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.

`-4.9t^2+1.5t+17=5`

My goal here in solving this equation is to get it into
`at^2+bt+c=0` so I can use the quadratic formula to solve it.

The quadratic formula is
`t=(-b \pm √(b^2-4ac))/(2a)`.

So let's begin that process here:

`-4.9t^2+1.5t+17=5`

Subtract 5 on both sides:

`-4.9t^2+1.5t+12=0`

So let's compare the following equations:

`-4.9t^2+1.5t+12=0`

`at^2+bt+c=0`.

`a=-4.9`

`b=1.5`

`c=12`

Now we are ready to insert in the quadratic formula:

`t=(-b \pm √(b^2-4ac))/(2a)`

`t=(-1.5 \pm √((1.5)^2-4(-4.9)(12)))/(2(-4.9))`

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
`(1.5)^2-4(-4.9)(12)`.

This gives you: 237.45

Let's show what we have so far now:

`t=(-b \pm √(b^2-4ac))/(2a)`

`t=(-1.5 \pm √((1.5)^2-4(-4.9)(12)))/(2(-4.9))`

`t=(-1.5 \pm √(237.45))/(2(-4.9))`

I'm going to put the denominator, 2(-4.9), into my calculator now.

`t=(-1.5 \pm √(237.45))/(-9.8)`

So this gives us two numbers to compute:

`t=(-1.5 - √(237.45))/(-9.8) \text{ and } t=(-1.5+√(237.45))/(-9.8)`

I'm actually going to type in -1.5-sqrt(237.45) into my calculator, as well as, -1.5+sqrt(237.45).

`t=(-16.90941271)/(-9.8) \text{ and } t=(13.90941271)/(-9.8)`

We are going to use the positive number only for our solution.

So we have the answer is whatever that first fraction is approximately:

`t=(-16.90941271)/(-9.8)=1.725450277`.

The answer is approximately 1.73 seconds.