# Speed of a Cliff Diver: Iraqi Olympic Diving

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## Introduction

This is a mechanics problem that a physics student should be able to solve.

Go to www.ebaumsworld.com/video/watch/119509/ and watch the Iraqi Olympic Diving video.

In the Iraqi diving video a man dives off a cliff into dust/earth/mud after which people clap and he receives a score for his dive.

if this video clip is analyzed the following can be determined:

1. Hang Time
2. Maximum Height
3. Height of Cliff
4. Vertical Takeoff Speed
5. Vertical Impact Speed

(Assume air resistance is negligible in all of the following circumstances)

## 1. Hang Time

Play the video clip in QuickTime. Count the frames in the man's diving trajectory. Distinguish between the diver's rising and falling portions.

total trajectory = 55 frames
rising = 7 frames
falling = 48 frames

Using QuickTime. go to "window", then click "show movie info". In the window were it says "movie fps" are the number of frames per second.

24 frames per second.

Now calculate the amount of time using

time = (number of frames)/(frames per second)
t(rising) = (7)/(24) = 0.29 seconds
t(falling) = (48)/(24) = 2.00 seconds
t(total trajectory) = (55)/(24) = (hang time) = 2.29seconds

## 2. Maximum Height

Use the amount of time of the diver falling to find the max height of the trajectory.

s = yo+ vot + (at2)/2
ymax = 0 + 0 + (9.81 m/s2)(2 seconds)2/2
ymax = 19.62 meters

## 3. Height of Cliff

Find the vertical distance the diver jumps

s = xo + vot + at2/2
s(vertical distance jumped) = 0 + 0 + (9.81 m/s2)(0.29)2/2
s(vertical distance jumped) = .19.62 meters

Subtract distance jumped from distance falling

(height of cliff) = (distance jumped)–(distance falling)
(height of cliff) = 19.62–.42 = 19.20 meters

## 4. Vertical Takeoff Speed

Use the distance jumped in the equation

vf2 = vo2 + 2as
vf = (vo2 + 2as)0.5
vf = (02 + 2 *9.82 *0.42).5 = 2.87 meters/second

## 5. Vertical Impact Speed

Use the distance falling in the equation

vf2 = vo2 + 2as
vf = (vo2 + 2as)0.5
vf = (02 + 2*9.82*19.62)0.5 = 19.62 m/s

Alexis Grisales -- 2004

Physics on Film