Question

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 1

2 revolutions

Step-by-step 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:

`s = gt^2/2`

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

`t^2 = (2s)/(g) = (2*8.3)/(9.8) = 1.694`

`t = √(1.694) = 1.3 s`

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

`rev = 1.3 * 1.6 = 2 revolution`