Download Earthquake, Tsunami and Nuclear Reactors

Professor Splash
11m into 30.5cm
http://www.youtube.com/watch?v=J54UzlFM9QU
How does he do it?
• He has to decrease his speed to zero in a short
amount of time
• This is done by interacting with the water and the
bottom of the pool.
• He ‘belly flops’ into the pool.
• If he landed feet first his feet would quickly go
through the water and hit the bottom.
Some of the (simple) Physics
• What energy transfer occurs as Prof Splash falls?
• GPE into KE
• Give the expression for the GPE transferred (KE
gained); keep the mass, m, in the expression.
• m x 9.8 x 11 (= 108m)
• How is this KE removed from Prof Splash?
• Work is done by the water.
• Give the expression for the work done by the water
• Fs (= F x 0.305)
• By equating work done to energy transferred write
an expression for the force applied by the water
• (F x 0.305) = (108m) so F = (108m) ÷ 0.305 = 354m
Some of the (harder) Physics
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Applying Newton’s 2nd Law
F – mg = ma so a = (F ÷ m) – (mg ÷ m)
a = 354m ÷ m – (mg ÷ m)
a = 354 – g
So the acceleration 344ms-2
Which means a g-force of 35g (or 35G)
F
mg
What have we not taken account of?
• More GPE is lost moving down through the water
– So more work has to be done by the water
– Hence more force is needed to stop Prof Splash
– So the acceleration will be more
• During the fall air resistance would mean not all
of the GPE is converted to KE
– Hence less force is needed to stop Prof Splash
• So the acceleration will be less
• The pool is resting on thick foam
– So Prof Splash will stop over a greater distance
– Hence less force is needed to stop Prof Splash
– So the acceleration will be less
Could be so much higher?
• Stopping from a greater height in the same water depth
will need a higher acceleration.
• The table gives tolerable accelerations from NASA data
• The equation is a quick way to calculate the acceleration
 height 
a  9.8
 1
 depth 
• “Eyeballs in” is the relevant column as this is the effect of
the acceleration on Prof Splash in the water because he
enters the water flat on his front.
How long does it take to stop?
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With a = 344ms-2 v = 0ms-1 s = 0.305m
s = vt - ½at2
t = √(2s/a) = √(2 x 0.305 ÷ 344)
The time is 0.42s
So Professor Splash is safe…at 11m