Impulse and maximum height of a particle

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thaalescosta
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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



[itex]\int(-8t²+8t)dt[/itex] = [itex](-8t³ +12t²)/3[/itex]

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.
 
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thaalescosta said:

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



[itex]\int(-8t^2+8t)dt[/itex] = [itex](-8t^3 +12t^2)/3[/itex]

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?
 
SammyS said:
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[itex]_{f}[/itex] - P[itex]_{i}[/itex] = ∫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[itex]_{max}[/itex] = v²/2g
then
h[itex]_{max}[/itex] = 9m²/32g


Does this make sense?
 
thaalescosta said:
Got it :)



So I = ΔP = P[itex]_{f}[/itex] - P[itex]_{i}[/itex] = ∫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[itex]_{max}[/itex] = v²/2g
then
h[itex]_{max}[/itex] = 9m²/32g

Does this make sense?
Yes.

You probably should enclose the entire denominator in parentheses.

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

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