Question

current attempt in progress
at the local swimming hole, a favorite trick is to run horizontally off a cliff that is 8.68 m above the water. one diver runs off the edge of the cliff, tucks into a \ball\, and rotates on the way down with an average angular speed of 2.37 revs. ignore air resistance and determine the number of revolutions she makes while on the way down.
number i units

Explanation:

Step1: Calculate the time of fall

The vertical motion is a free - fall. Using the equation \(h = v_{0y}t+\frac{1}{2}gt^{2}\), since \(v_{0y} = 0\) (horizontal run), we have \(h=\frac{1}{2}gt^{2}\).
Solving for \(t\), we get \(t=\sqrt{\frac{2h}{g}}\).
Substituting \(h = 8.68\ m\) and \(g = 9.8\ m/s^{2}\), we have \(t=\sqrt{\frac{2\times8.68}{9.8}}\).

$$t=\sqrt{\frac{17.36}{9.8}}\approx\sqrt{1.7714}\approx1.33\ s$$

Step2: Calculate the number of revolutions

The angular speed \(\omega=2.37\ rev/s\).
The number of revolutions \(N=\omega t\).
Substituting \(\omega = 2.37\ rev/s\) and \(t = 1.33\ s\), we get \(N=(2.37)(1.33)\).

$$N = 3.1521\approx3.15$$

Answer:

\(3.15\) revolutions