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

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In a discussion about a juggler performing under a 2-meter ceiling, the maximum upward speed of the ball is calculated to be 6.26 m/s to prevent it from hitting the ceiling. The time the ball remains in the air before being caught is approximately 1.25 seconds, derived from the equation of motion. The calculations involve using gravitational acceleration and the maximum height to determine the time. The thread highlights the importance of understanding basic physics principles in solving such problems. Overall, the discussion emphasizes straightforward methods for solving projectile motion in a confined space.
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$\textsf{A juggler performs in a room with a ceiling 2 m above hand level.}$
$\textit{a. what is the maximum upward speed she can give a ball }$
$\textit{without letting the ball hit the ceiling.$\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
 
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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:

$$t\approx2\sqrt{\frac{2(2\text{ m})}{9.8\frac{\text{m}}{\text{s}^2}}}\approx1.28\text{ s}$$
 
I should always come here first
saw some other examples
but they didn't make sense
and was a lot more complicated
 
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