Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 8 - Rotational Kinematics - Problems - Page 214: 31


The number of revolutions the diver makes is $2.08$ revolutions.

Work Step by Step

We assume the diver, at the time when she jumps off the cliff, has zero vertical speed, or $v_0=0$. Combined with the information about the cliff's height, $s=8.3m$ and gravitational acceleration $g=9.8m/s^2$, we can calculate the time she is in the air: $$s=v_0t+\frac{1}{2}gt^2=0+\frac{1}{2}gt^2$$ $$t=\sqrt{\frac{2s}{g}}=1.301s$$ During this time, she has $\overline\omega=1.6rev/s$. The number of revolutions she makes is $$\theta=\overline\omega t=2.08rev$$
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