How do you calculate the impulse on a person performing a belly flop?

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Homework Help Overview

The problem involves calculating the impulse experienced by a person performing a belly flop from a height of 12 meters into water, with the assumption that they stop upon reaching the bottom of the water. The discussion centers around the concepts of impulse, momentum, and the relevant equations of motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of final velocity using kinematic equations and the implications of estimating mass. There is a focus on understanding the relationship between initial and final momentum and how impulse is defined in this context.

Discussion Status

Participants are actively questioning the assumptions made regarding the height of the drop and the calculation of impulse. Some have provided calculations and observations, while others are clarifying definitions and the conditions under which impulse is measured. There is no explicit consensus yet, but the discussion is progressing with various interpretations being explored.

Contextual Notes

There is uncertainty regarding the correct interpretation of the height from which the impulse is calculated, as well as the sign convention used in the calculations. Participants are also considering the implications of the estimated mass on the final results.

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Homework Statement


Until his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 12 m into 30 cm of water. Assuming that he stops just as he reaches the bottom of the water and estimating his mass, find the magnitude of the impulse on him from the water.


Homework Equations


v^2 = vo^2 +2ay
j = pf - pi
p = mv


The Attempt at a Solution


y = 12.3
a = 9.8

from equation 1, Vf = 15.53 m/s

from equation 2 and 3 J = 15.53m -0m = 15.53m

Help please.
 
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Shatzkinator said:

Homework Statement


Until his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 12 m into 30 cm of water. Assuming that he stops just as he reaches the bottom of the water and estimating his mass, find the magnitude of the impulse on him from the water.

Homework Equations


v^2 = vo^2 +2ay
j = pf - pi
p = mv

The Attempt at a Solution


y = 12.3
a = 9.8

from equation 1, Vf = 15.53 m/s

from equation 2 and 3 J = 15.53m -0m = 15.53m

Help please.

What is his mass or what is it you estimate his mass to be?

You've calculated the Δ of velocity, but j = Δp

One small observation though. The Δp is taken as he contacts the water to at the bottom? That is where is p_final is taken. So wouldn't his initial v be based on a drop of 12m, not 12.3?
 
LowlyPion said:
What is his mass or what is it you estimate his mass to be?

You've calculated the Δ of velocity, but j = Δp

One small observation though. The Δp is taken as he contacts the water to at the bottom? That is where is p_final is taken. So wouldn't his initial v be based on a drop of 12m, not 12.3?

I'm not sure what you mean by the second part of your comment. For Δp, v1 = 0, vf = like you said at the bottom. So the total height he travels is 12 + 0.3 from depth of water. I don't understand what you're trying to say.
 
Shatzkinator said:
I'm not sure what you mean by the second part of your comment. For Δp, v1 = 0, vf = like you said at the bottom. So the total height he travels is 12 + 0.3 from depth of water. I don't understand what you're trying to say.

That is the total distance of travel. But doesn't the question ask what is the impulse from the water. Wouldn't that be measured from the initial momentum at 12m? to the 0 velocity at the bottom?
 
LowlyPion said:
That is the total distance of travel. But doesn't the question ask what is the impulse from the water. Wouldn't that be measured from the initial momentum at 12m? to the 0 velocity at the bottom?

Okay but now I get Δp = 0 - 15.33(65) where 65 is estimated mass
= -996.9 Ns

The book shows the answer as 1.0 x 10^3 to 1.2 x 10^3... positive values. =S
 
Shatzkinator said:
Okay but now I get Δp = 0 - 15.33(65) where 65 is estimated mass
= -996.9 Ns

The book shows the answer as 1.0 x 10^3 to 1.2 x 10^3... positive values. =S

What is positive y?

It's not (0 - (-15.33)) ?
 
LowlyPion said:
What is positive y?

It's not (0 - (-15.33)) ?

ahh okay, i had the acceleration and y both as positive!

thank you! =)
 

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