Question

at the local swimming hole, a favorite trick is to run horizontally off a cliff that is 5.44 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 1.87 rev/s. ignore air resistance and determine the number of revolutions she makes while on the way down.

Explanation:

Step1: Find the time of fall

Use the equation \(h = v_{0y}t+\frac{1}{2}gt^{2}\). Since \(v_{0y} = 0\) (horizontal run), \(h=\frac{1}{2}gt^{2}\). Solving for \(t\):
\(t=\sqrt{\frac{2h}{g}}\)
Substitute \(h = 5.44\space m\) and \(g = 9.8\space m/s^{2}\)
\(t=\sqrt{\frac{2\times5.44}{9.8}}=\sqrt{\frac{10.88}{9.8}}\approx1.05\space s\)

Step2: Calculate the number of revolutions

Use the formula \(\theta=\omega t\), where \(\omega = 1.87\space rev/s\) and \(t = 1.05\space s\)
\(\theta=(1.87\space rev/s)\times(1.05\space s)\approx1.96\space rev\)

Answer:

\(1.96\) revolutions