# Impulse and maximum height of a particle

1. Feb 16, 2014

### thaalescosta

1. The problem statement, all variables and given/known data
A particle receives an impulse that lasts 1s, coming from a upwards vertical force. This force is given by the following equation: F = -8t²+8t.

What is the maximum height reached by the particle?

2. Relevant equations

F = -8t²+8t

3. The attempt at a solution

$\int(-8t²+8t)dt$ = $(-8t³ +12t²)/3$

When t = 1s, I = 4/3. That's what I got as my impulse.

For the maximum height, I took the derivative of F and set it = 0, finding that t = 1/2

Now I don't know how to find the height with all the information I have.

2. Feb 16, 2014

### SammyS

Staff Emeritus
Hello thaalescosta. Welcome to PF !

(Use ^2 for an eponent of 2 in Latex.)

There's no need to use the derivative of the force, F.

The change in momentum of the particle is equal to the impulse. (Impulse - Momentum Theorem)

Do you know the mass of the particle?

3. Feb 16, 2014

### thaalescosta

Got it :)

So I = ΔP = P$_{f}$ - P$_{i}$ = ∫Fdt, from i to f

My impulse turned out to be 4/3.

The mass of the particle wasn't given. I'm guessing that the answer will be in general terms.

If I = m.v, then v = 3m/4

And if the maximum height is
h$_{max}$ = v²/2g
then
h$_{max}$ = 9m²/32g

Does this make sense?

4. Feb 16, 2014

### SammyS

Staff Emeritus
Yes.

You probably should enclose the entire denominator in parentheses.

$h_{max} = v^2/(2g)$

$h_{max} = 9m^2/(32g)$