Question

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is 8.0 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 1.2 rev/s. Ignore air resistance and determine the number of revolutions she makes while on the way down.

Answer 1

Answer:

Number of revolutions=1.532 revolutions

Explanation:

Given data

Distance s=8.0 m

Angular speed a=1.2 rev/s

To find

Number of revolutions

Solution

From the equation of simple motion we not that

$S=ut+1/2gt^{2}\\ where\\u=0\\So\\8.0m=0+(1/2)(9.8m/s^{2} )t^{2}\\ t^{2}=\frac{8.0m}{0.5*9.8m/s^{2} } \\ t^{2}=1.63\\t=\sqrt{1.63} \\t=1.28s$

So for the number of revolutions she makes is given as

$n=a*t\\n=(1.2rev/s)(1.28s)\\n=1.532revolutions$