Solving Impulse on Jumper for 50 kg Woman

[Mr Davis 97]

Homework Statement

A 50 kg woman jumps straight into the air, rising h meters from the ground. What impulse does she receive from the ground to attain this height?

Homework Equations

The Attempt at a Solution

Impulse is defined as a change in momentum. Thus, $I_{woman} = m(v_f - v_i)$. Since she jumps to some height, we know that her final velocity is zero. Thus$I_{woman} = - m v_i$. We can find the initial velocity with $v_f^2 = v_i^2 - 2gh$, so $v_i = \sqrt{2gh}$. Thus, the impulse on the woman from the ground is $I_{woman} = -m \sqrt{2gh}$.

So the magnitude of this impulse is correct, but apparently my sign is wrong. What am I doing wrong?

 

[drvrm]

Impulse is defined as a change in momentum. Thus, Iwoman=m(vf−vi)Iwoman=m(vf−vi)I_{woman} = m(v_f - v_i). Since she jumps to some height,

one defines impulse as forcex differential time element which is equal to change in momentum as force is rate of change in momentum

therefore the lady had a change in momentum from rest to massx velocity of projection;
if somebody is taking the full time to reach height h in that situation one solves the equation of motion with gravitational force(and can get the velocity of projection) and 'impulse' imparted to the jumping lady is not being defined.

 

[Mr Davis 97]

I don't really understand what you mean... What's the specific thing that I am doing wrong in terms of the sign of the impulse?

 

[drvrm]

What's the specific thing that I am doing wrong in terms of the sign of the impulse?

Impulse = massx Initial velocity of Jumper - momentum at rest(which is zero) = change in momentum

 

[Mr Davis 97]

I thought that impulse was just the final momentum minus the initial momentum?

 

[drvrm]

I thought that impulse was just the final momentum minus the initial momentum?

Impulsive forces are short duration action on a body therefore it is defined as F.delta(T)

as F = d (m.v)/dt
one comes to the impulse as change in momentum during contact.

 

[Mr Davis 97]

So your saying that the change in momentum is all during contact with the ground, which means that the initial velocity is zero, and the final velocity is the velocity at which she leaps from the ground?

 

[PeroK]

So your saying that the change in momentum is all during contact with the ground, which means that the initial velocity is zero, and the final velocity is the velocity at which she leaps from the ground?

She receives a positive impulse from the ground that gets her moving. But, all that momentum is lost to gravity by the time she reaches the highest point (this is what you calculated).

You should have used the projectile equations to calculate the velocity $v$ with which she left the ground, but this is the final velocity with respect to the impulse from the ground.

In general, you have to think more carefully about the stage of the overall motion you are considering. In this problem the final velocity for stage 1 (pushing off from the ground) becomes the initial velocity for stage 2 (the projectile/free fall stage).

 

[drvrm]

So your saying that the change in momentum is all during contact with the ground, which means that the initial velocity is zero, and the final velocity is the velocity at which she leaps from the ground?

thanks @PeroK
Mr. Davis Another good example of Impulse is when you throw a ball towards a wall and the ball returns taking some path like a projectile during impact the momentum changes and impulse can be calculated by change in momentum .
here if you try to jump from floor you push the surface beneath your feet and get a reaction force which projects you with a velocity and the impulse is change in momentum equals to final momentum as initial momentum is zero.
i think now you could get the picture.

 

[haruspex]

Impulsive forces are short duration action on a body therefore it is defined as F.delta(T)

No, that's too restrictive. Impulse can be spread over any length of time. But we often use the concept of impulse because it can be applied in the case of arbitrarily large forces, with unknown time profiles, acting over arbitrarily short periods of time.

So your saying that the change in momentum is all during contact with the ground, which means that the initial velocity is zero, and the final velocity is the velocity at which she leaps from the ground?

Yes, the question is asking about impulse from the ground leading up to the moment of losing contact with it. However, the question is not quite right. The complete impulse from the ground over that time alse needs a component to counteract gravity until losing contact. If it took some short time t from starting the leap to loss of contact, the total impulse is m√(2gh)+mgt.