Question

Question #d8707

Answer 1

`sqrt30 " "or" " 5.477`

Explanation:

We are given that the height, `f(x)`, is equal to 50.

The problem wants us to find the time, `x`, at which the height is 50m.

Therefore, we can plug in `color(red)50` for `f(x)`, and solve for `x`.

`f(x) = -5x^2 + 200`

`color(red)50 = -5x^2 + 200`

`color(red)50 - color(blue)200 = -5x^2 + cancel200 - cancelcolor(blue)200`

`-150 = -5x^2`

`(-150)/color(blue)(-5) = (cancel(-5)x^2)/cancelcolor(blue)(-5)`

`30 = x^2`

`color(blue)(sqrtcolor(black)(30)) = color(blue)(sqrtcolor(black)(x^cancel2)`

`sqrt30 = x`

So the ball will be 50 meters above the ground at `sqrt30`, or about `5.477` seconds.

Final Answer