Question d8707

Jan 27, 2018

$\sqrt{30} \text{ "or" } 5.477$

Explanation:

We are given that the height, $f \left(x\right)$, is equal to 50.

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

Therefore, we can plug in $\textcolor{red}{50}$ for $f \left(x\right)$, and solve for $x$.

$f \left(x\right) = - 5 {x}^{2} + 200$

$\textcolor{red}{50} = - 5 {x}^{2} + 200$

$\textcolor{red}{50} - \textcolor{b l u e}{200} = - 5 {x}^{2} + \cancel{200} - \cancel{\textcolor{b l u e}{200}}$

$- 150 = - 5 {x}^{2}$

$\frac{- 150}{\textcolor{b l u e}{- 5}} = \frac{\cancel{- 5} {x}^{2}}{\cancel{\textcolor{b l u e}{- 5}}}$

$30 = {x}^{2}$

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

$\sqrt{30} = x$

So the ball will be 50 meters above the ground at $\sqrt{30}$, or about $5.477$ seconds.