Uniform accelerated motion (UAM) equations:

$\overline{)\mathbf{}{{\mathit{v}}}_{{\mathit{f}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathbf{}}{\mathbf{+}}{\mathit{a}}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{x}}{\mathbf{=}}{\mathbf{}}\mathbf{\left(}\frac{{\mathit{v}}_{\mathit{f}}\mathbf{+}{\mathit{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{x}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathit{t}}{\mathbf{+}}{\frac{1}{2}}{\mathit{a}}{{\mathit{t}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{{{\mathit{v}}}_{{\mathit{f}}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{}}{{{\mathit{v}}}_{{\mathbf{0}}}}^{{\mathbf{2}}}{\mathbf{}}{\mathbf{+}}{\mathbf{2}}{\mathit{a}}{\mathbf{\u2206}}{\mathit{x}}}$

The blue ball reaches 0.90m at frame 1181.

t = (change in frames)/(frames/s) = (1181 - 130)/2400 = 0.438 s

Δx = v_{0}t + (1/2)at^{2}

Video text description for the Direct Measurement Video of Falling Through the Air

In this video, three different balls are held above the ground by ribbons: a blue foam ball with mass 3.15 grams; a small orange steel ball with mass 67.1 grams; and a large purple bowling ball with mass 5539 grams. A vertical scale, marked off in centimeters, is aligned with each ball, with the bottom of each ball at the 0-centimeter line at the beginning of the video. The video is shot at 2400 frames per second, so plays at 80 times slower than normal speed. A frame counter indicates the current frame.

At frame 130, someone cuts each ribbon at exactly the same time and the balls all start falling under the influence of gravity. At first, all the balls seem to move together, but it soon becomes clear that the blue ball takes longer to go the same distance as the others. The orange and purple balls both reach the 10-centimeter mark at around frame 472, while the blue ball takes two frames longer, reaching the 10-centimeter mark at frame 474. Later, the orange and purple balls both reach the 90-centimeter mark at around frame 1157, while the blue ball takes all the way until frame 1181 to reach 90 centimeters.

Use the same approach as above to find the average acceleration of the blue ball as it falls through 0.90 m. Remember that the motion is assumed to start at frame 130. Express your answer in m/s^{2} with three significant figures.

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