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

In summary, the question asks to find the magnitude of the impulse on Henri LaMothe as he belly-flops into 30 cm of water from a height of 12 m. Using the given equations and an estimated mass, the final velocity is calculated to be 15.53 m/s. The magnitude of the impulse is then found to be 996.9 Ns, with the correct answer being 1.0 x 10^3 to 1.2 x 10^3 Ns. The mistake was made by using the wrong sign for the acceleration and distance variables.
  • #1
Shatzkinator
53
0

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.
 
Physics news on Phys.org
  • #2
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?
 
  • #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.
 
  • #4
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?
 
  • #5
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
 
  • #6
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)) ?
 
  • #7
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! =)
 

1. What is a collision?

A collision is an event where two or more objects come into contact with each other, resulting in a change in their motion or shape.

2. How is the momentum of objects affected by a collision?

In a collision, the total momentum of the system remains constant. This means that the momentum of the objects involved may change, but the sum of their momentums before and after the collision will be the same.

3. What is impulse and how is it related to collisions?

Impulse is the change in momentum of an object. In a collision, the impulse experienced by an object is equal to the force applied to it multiplied by the time over which the force acts.

4. What factors affect the amount of force and impulse in a collision?

The amount of force and impulse in a collision can be affected by the mass, velocity, and angle of impact of the objects involved, as well as the duration of the collision and the materials the objects are made of.

5. How are collisions and impulse important in real-world applications?

Collisions and impulse are important in many real-world applications, such as car crashes, sports, and engineering. Understanding these concepts can help predict and prevent accidents, improve athletic performance, and design safer and more efficient structures and machines.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
6K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
886
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
Back
Top