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A diver makes 2.5 revolution on the way from a 10-m-high platform to the water. Assuming zero initial vertical velocity, the average angular velocity during the dive is

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$\frac{3 π}{\sqrt{2}} rad / s$

$\frac{5 π}{\sqrt{2}} rad / s$

$\frac{5 π}{\sqrt{3}} rad / s$

$\frac{π}{\sqrt{2}} rad / s$

answer is B.

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Detailed Solution

The free-fall time:

$∆ y = v_{o} t + \frac{1}{2} {gt}^{2} \Rightarrow t = \sqrt{\frac{2 (10 m)}{10 m / s^{2}}} = \sqrt{2} s$

Thus, the magnitude of the average angular velocity is

$ω_{avg} = \frac{(2.5 rev) (2 π rad / dev)}{\sqrt{2} s} = \frac{5 π}{\sqrt{2}} rad / s$

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