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.

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

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

 

[HalfLostAndSearching]

I would approach this problem by first defining some variables and assumptions. Let's say that the particle has a mass of m and starts at a height of h=0. We can also assume that the force acting on the particle is due to gravity, so we can use the equation F=mg. From the given equation, we can see that the force varies with time, so we can use the impulse-momentum theorem to calculate the change in momentum of the particle.

Impulse is defined as the change in momentum, so we can write the equation as:

I = m(vf-vi)

Where I is the impulse, m is the mass of the particle, vf is the final velocity, and vi is the initial velocity. Since the particle starts at rest, vi=0 and we can rewrite the equation as:

I = mvf

From the given equation, we can see that the force acting on the particle is given by F=-8t^2+8t. We can use the equation F=ma to relate the force to the acceleration of the particle. Substituting this into the impulse-momentum theorem, we get:

I = m(vf-vi) = m∫F dt

We can solve this integral using the limits of integration from t=0 to t=1s, since the impulse only acts for 1 second. This gives us:

I = m∫(-8t^2+8t) dt = m(-8t^3/3 + 4t^2)

Evaluating this at t=1s, we get I=4/3m.

Now, we can use this value of impulse and the initial velocity (vi=0) to calculate the final velocity of the particle:

I = mvf, so vf = I/m = (4/3m)/m = 4/3 m/s

To find the maximum height reached by the particle, we can use the equation for displacement under constant acceleration:

Δy = viΔt + 1/2a(Δt)^2

Since the particle starts at rest, vi=0 and the equation simplifies to:

Δy = 1/2a(Δt)^2

Substituting in the values for acceleration (a=-8t+8) and time (Δt=1s), we get:

Δy = 1/2