# Rotational Kinematics of Running

1. Apr 24, 2006

### MetalCut

Hi
I need some help.
The question is:

At the local swimming hole, a favourite trick is to run horizontally of a cliff that is 8.3m 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 1.6rev/s. Ignore air resistance and determine the number of revolutions she makes on the way down.

Now i've done it like this but the anser is wrong.........

Well, gravity pulls her down at 9.8m/s². So in the first second she will fall 9.8m.
She only needs to fall 8.3m to the water.
So 8.3m ÷ 9.8m = 0.85s
It will take her only 0.85s to reach the water.
And she makes 1.6rev/s.
So 1.6 ÷ 100 = 0.016
Then 0.016 x 85 = 1.36
Thus she will make 1.36 revolutions on her way down.

What am i missing?

2. Apr 24, 2006

### Harmony

9.8ms^-2 means that the body will achieve the velocity of 9.81ms^-1 in a second, not 9.81m.
I think the equation, s=ut+0.5at^2 should be used instead.

3. Apr 24, 2006

### HallsofIvy

Staff Emeritus
You are assuming her speed was a constant 9.8 m/s which is not true.
As Harmony said, use s= ut+ 0.5at^2. Here u (initial speed) is 0 and a (acceleration) is -9.8. Since she is diving down 8.3 m, -8.3= -(0.5)(9.8)t^2.

4. Apr 24, 2006

### MetalCut

Thanx alot, i dont know how i could have made such a stupid mistake......

Thanx