Question

A diver makes $2.5$ revolutions 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

• A. $\frac{3 \pi}{\sqrt{2}} rad / s$
• B. $\frac{5 \pi}{\sqrt{2}} rad / s$
• C. $\frac{5 \pi}{\sqrt{3}} rad / s$
• D. $\frac{\pi}{\sqrt{2}} rad / s$

Answer 1

Answer: (B)

The free-fall time:
$\Delta y=v_{0 y} t+\frac{1}{2} g t^{2}$
$ \Rightarrow t=\sqrt{\frac{2(10 \,m )}{10\, m / s ^{2}}}=\sqrt{2} s$
Thus, the magnitude of the average angular velocity is
$\omega_{ avg }=\frac{(2.5\, rev )(2 \pi \,rad / rev )}{\sqrt{2} s }=\frac{5 \pi}{\sqrt{2}} rad / s$