# IMPULSE, i need HELP

1. Nov 8, 2007

### physicsbhelp

[SOLVED] IMPULSE, i need HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1. The problem statement, all variables and given/known data

Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 11 m into 30 cm. of water (Figure 9-49). Assuming that he stops just as he reaches the bottom of the water and estimating his mass to be 76 kg, find the magnitudes of the impulse on him from the water.

2. Relevant equations

Impulse = Force * Time
Ug = mgh

3. The attempt at a solution

Ug = (76)(9.81)(11-0.3) = 7,977.492

2. Nov 8, 2007

### physicsbhelp

1. The problem statement, all variables and given/known data

Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 11 m into 30 cm. of water (Figure 9-49). Assuming that he stops just as he reaches the bottom of the water and estimating his mass to be 76 kg, find the magnitudes of the impulse on him from the water.

2. Relevant equations

Impulse = Force * Time
Ug = mgh

3. The attempt at a solution

Ug = (76)(9.81)(11-0.3) = 7,977.492

DO I USE PROJECTILE MOTION, and if i do, then how?

3. Nov 8, 2007

4. Nov 8, 2007

### aq1q

ok well.. thats not the only formula... its also impulse = mv1-mv2

5. Nov 9, 2007

### Staff: Mentor

To solve the problem, one needs to find the velocity just as LaMothe hits the water.

Equate the change in gravitational potential energy (mgh), h = 76 m, with the change in KE = 1/2 mv2, and use that to get v at the time LaMothe impacts the water.

From that, one gets mv1. LaMothe must competely stop in 0.3 m of water, so determine the average velocity at constant deceleration. Then compute the time interval to stop by 0.3 m/(vav).

Here is a good reference - http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#mot1

Impulse - http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html#c1

6. Nov 9, 2007

### aq1q

If you haven't learned work energy theorem, use basic kinematic formulas such as V(final)^2=V(initial)^2-2g(deltaX).

Now, I could be wrong, but i think thats it... your v2=0, so u mV1 is the impulse.. now if u were asked to find the Fmax, or the force at any given time while its in the water, then its different.

Last edited: Nov 9, 2007
7. Nov 9, 2007

### aq1q

also we dont have a figure; so, there could some minor chance that I didn't understand the problem completely

8. Nov 9, 2007

### Feldoh

He still moves a distance of 30 cm after he hits the water, so v2 wouldn't be 0.

9. Nov 9, 2007

### Staff: Mentor

That is correct with mv1 being the momentum at contact with the water and v= 0 (mv = 0) at the bottom of the water.

Then I = m$\Delta$v = m(V1-0) = mv1.