Question

A rock is thrown from the side of a ledge. The height (in feet), x seconds after the toss is modeled by: h(x) = -3(x-4)^2+79 How many seconds pass before the rock lands on the ground? Round your answer to the nearest tenths place.

Answer 1

The rock lands on the graound when h(x) = 0:

`h(x)=-3(x-4)^2+79=0`
`-3(x-4)^2=-79`

Dividing both sides by -3 gives

`(x-4)^2=(79)/(3)`

Taking the square root of both sides gives

`x-4=\sqrt[]{(79)/(3)}`

Finally, adding 4 to both sides gives

`x=\sqrt[]{(79)/(3)}+4`

converting to a decimal form gives

`x=9.1`

Hence, the rock lands on the ground after 9.1 s.