Impulse and maximum height of a particle

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Homework Help Overview

The discussion revolves around a particle subjected to an upward vertical force described by the equation F = -8t² + 8t, receiving an impulse over a duration of 1 second. Participants are exploring how to determine the maximum height reached by the particle based on the impulse and force provided.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of impulse and its relationship to momentum. There is an exploration of how to derive the maximum height from the impulse without knowing the mass of the particle. Questions arise regarding the necessity of using the derivative of the force and the implications of missing mass information.

Discussion Status

Some participants have provided guidance on the relationship between impulse and momentum, suggesting that the maximum height can be expressed in general terms due to the unknown mass. There is an ongoing exploration of the implications of the calculations presented, but no consensus has been reached on the final interpretation of the results.

Contextual Notes

The mass of the particle is not provided, which affects the ability to calculate specific numerical results for the maximum height. Participants are considering the problem in general terms as a result.

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



\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.
 
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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



\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?
 
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_{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?
 
thaalescosta said:
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)
 

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