MHB Cp2.27 How long is the ball in the air before it is caught

[karush]

$\textsf{A juggler performs in a room with a ceiling 2 m above hand level.}$
$\textit{a. what is the maximum upward speed she can give a ball }$
$\textit{without letting the ball hit the ceiling.$\displaystyle 6.26 \frac{m}{s}$}$
\begin{align*}
v^2&=2(9.8)(2)\\
&=39.2\\
v&=\sqrt{39.2}\\
&=\color{red}{6.26 \, \frac{m}{s}}
\end{align*}
$\textit{b. How long is the ball in the air before it is caught}$
answer is $1.25 s$ok I couldn't figure out b.
altho its simple

 

[MarkFL]

I would use:

$$x=-\frac{1}{2}gt^2+v_0t=0$$

Take the non-zero root, so solve:

$$-\frac{1}{2}gt+v_0=0$$

$$t=\frac{2v_0}{g}=\frac{2\sqrt{2gx_{\max}}}{g}=2\sqrt{\frac{2x_{\max}}{g}}$$

Substitute for the given data:

$$t\approx2\sqrt{\frac{2(2\text{ m})}{9.8\frac{\text{m}}{\text{s}^2}}}\approx1.28\text{ s}$$

 

[karush]

I should always come here first
saw some other examples
but they didn't make sense
and was a lot more complicated