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Problem: At the local swimming hole, a favorite trick is torun
horizontally off a cliff that is 8.3m above the water. Once diver
runs off the edge of the cliff, tucks into a ball androtates on the
way down with an average angular speed of 1.6rev/s.
Ignore air resistance and determine the number of revolutions
shemakes while on the way down.

Answer :

MechEngineer

Answer:

2 revolutions

Explanation:

Assume that when she runs off the edge of the 8.3m high cliff, her vertical speed is 0. So gravitational acceleration g = 9.8m/s2 is the only thing that makes her fall down. So we can use the following equation of motion to calculate the time it takes for her to fall down:

[tex]s = gt^2/2[/tex]

where s = 8.3 m is the distance that she falls, t is the time it takes to fall, which is what we are looking for

[tex]t^2 = \frac{2s}{g} = \frac{2*8.3}{9.8} = 1.694 [/tex]

[tex]t = \sqrt{1.694} = 1.3 s[/tex]

Since she rotates with an average angular speed of 1.6rev/s. The number of revolutions she would make within 1.3s is

[tex]rev = 1.3 * 1.6 = 2 revolution[/tex]

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