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

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  • Thread starter karush
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In summary, a juggler performs in a room with a ceiling 2 m above hand level. The maximum upward speed she can give a ball without letting it hit the ceiling is 6.26 m/s. The ball is in the air for approximately 1.25 seconds before it is caught. To calculate the time the ball is in the air, the equation $x=-\frac{1}{2}gt^2+v_0t=0$ can be used, where $g$ is the acceleration due to gravity and $v_0$ is the initial upward speed. The non-zero root of this equation is given by $t=\frac{2v_0}{g}$, which for the given data
  • #1
karush
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MHB
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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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  • #2
I would use:

\(\displaystyle x=-\frac{1}{2}gt^2+v_0t=0\)

Take the non-zero root, so solve:

\(\displaystyle -\frac{1}{2}gt+v_0=0\)

\(\displaystyle t=\frac{2v_0}{g}=\frac{2\sqrt{2gx_{\max}}}{g}=2\sqrt{\frac{2x_{\max}}{g}}\)

Substitute for the given data:

\(\displaystyle t\approx2\sqrt{\frac{2(2\text{ m})}{9.8\frac{\text{m}}{\text{s}^2}}}\approx1.28\text{ s}\)
 
  • #3
I should always come here first
saw some other examples
but they didn't make sense
and was a lot more complicated
 

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

1) How is the time in the air calculated for a falling object?

The time in the air for a falling object can be calculated using the equation t = √(2h/g), where t is the time in seconds, h is the height in meters, and g is the acceleration due to gravity (9.8 m/s²).

2) Is the time in the air affected by air resistance?

Yes, air resistance can affect the time in the air for a falling object. The more air resistance there is, the longer the object will take to fall and therefore the longer it will be in the air before it is caught.

3) What factors can affect the time in the air for a thrown ball?

The time in the air for a thrown ball can be affected by the initial velocity, angle of release, air resistance, and the height at which it is thrown.

4) How does the height at which the ball is thrown affect the time in the air?

The higher the ball is thrown, the longer it will take to reach the ground and therefore the longer it will be in the air before it is caught.

5) Can the time in the air be calculated for an object thrown at an angle?

Yes, the time in the air for an object thrown at an angle can be calculated using the equation t = (2v₀sinθ)/g, where t is the time in seconds, v₀ is the initial velocity, θ is the angle of release, and g is the acceleration due to gravity.

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