Question

Question #6c732

Answer 1

Sorry, what's the question here? 6 seconds, if you're looking for the time it takes for the rock to hit the ground.

Explanation:

It's not super clear what you're asking (no offence).

If you're asking how long it takes the rock to hit the ground (just my guess) I can answer that.

We know the height is given as
`h(t)=-16x^2+16x+480`

though I feel like the "x"s were supposed to be "t"s, since h was a function of t and you didn't mention t in the equation. To rewrite,

`h(t)=-16t^2+16t+480`

We also know that when the rock hits the ground,
`h(t)=0`.

So we need to solve the quadratic for when `h(t)=0`

`0=-16t^2+16t+480`
We can divide by 16
`0/16=-t^2+t+30`
`0=-t^2+t+30`

Now we can factor this or use the quadratic formula . Just for fun, let's use the quadratic formula. I'm assuming you know how to solve quadratics, but if not perhaps check out the links above.

`t=\frac{-1 \+- \sqrt(1^2-4(-1)(30))}{2(-1)}`
`=\frac{-1 \+- \sqrt(121)}{-2}`
`=\frac{1 \+- \sqrt(121)}{2}`
`=\frac{1 \+- 11}{2}`
`=\frac{12}{2} or -10/2`
`=6 or -5`
But the negative 5 doesn't really make sense in this case (the equation doesn't apply for t less than 0), so let's take the value of 6.

So, after 6 seconds, the rock will hit the ground.