# Impulse and maximum height of a particle

• thaalescosta
In summary, the maximum height reached by a particle after receiving an impulse lasting 1 second from an upwards vertical force, with an equation of F = -8t^2+8t, is 9m^2/(32g). The mass of the particle was not given, so the answer is in general terms.
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?

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.

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?

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.

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

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

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

## 1. What is impulse?

Impulse is the change in momentum of an object, calculated as the force applied to the object multiplied by the time interval over which the force is applied.

## 2. How is the impulse-momentum theorem related to impulse?

The impulse-momentum theorem states that the impulse applied to an object is equal to the change in momentum of the object. This means that the larger the impulse, the greater the change in momentum and therefore the greater the change in the object's speed or direction.

## 3. How does impulse affect the maximum height of a particle?

Impulse plays a crucial role in determining the maximum height that a particle can reach. The greater the impulse applied to the particle, the greater the change in its momentum and the higher it will be able to reach.

## 4. What factors can affect the impulse and maximum height of a particle?

The impulse and maximum height of a particle can be affected by factors such as the initial velocity of the particle, the angle of projection, and the force applied to the particle. Other environmental factors, such as air resistance, can also influence these values.

## 5. How can the impulse and maximum height of a particle be calculated?

The impulse and maximum height of a particle can be calculated using the equations of motion and the principles of conservation of momentum. These calculations may be more complex for objects with varying forces acting on them, but can be simplified for objects with constant forces and motion.

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