Impulse and maximum height of a particle

[thaalescosta]

Homework Statement

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?

Homework Equations

F = -8t²+8t

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.

 

[SammyS]

Homework Statement

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?

Homework Equations

F = -8t²+8t

The Attempt at a Solution

\int(-8t^2+8t)dt = (-8t^3 +12t^2)/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.

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?

 

[thaalescosta]

Hello thaalescosta. Welcome to PF !

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

Got it :)

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?

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?

 

[SammyS]

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?

Yes.

You probably should enclose the entire denominator in parentheses.

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

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