Impulse required to move an object vertically against gravity to height $z$ assuming zero initial velocity, zero final velocity and zero drag force The work done $E$ to move an object of weight $W$, directly against gravity by vertical distance $z$ is simply:
$E= W z$
Now, I have found the following link How to calculate impulse required to move an object vertically upward by given distance  which gives the impulse required (even with drag force) but assumes that the object has an initial velocity.
My question is: provided that we ignore drag forces and assume that the object starts with zero velocity and ends with zero velocity (and thus the object only has to overcome its own weight), should there not be a required impulse to cause the object to move this distance against gravity? If so, how do you prove this and what is the answer?
I realise that if you assume it has an initial velocity and thereafter no work is done (and finishes with zero velocity), the solution is simply:
$J=(2mWz)^{0.5}$
Where $J$ is the impulse imparted and $m$ is the mass of the object. This is found by equating the kinetic and potential energies.
I'm starting to think that there isn't a ubiquitous equation and it depends on the amount of time over which it is imparted?
Some help here would be greatly appreciated!
 A: I think you got it right. If you have a body with mass $m$ at rest and you like to lift it to the height $h$ against gravity, you need to provide the energy $E=mgh$. Hence, you need to apply a force $F$ over a distance $s$, which satisfies 
$$
m g (h-s) = E = F\cdot s
= m a \cdot s
$$
Here we assumed that $F=const$. Furthermore, if we assume that the distance $s$ over which the force is applied is much smaller that the height $h$, we obtain the relation
$$
gh = a \cdot s = a \cdot \frac{1}{2}at^2
\Rightarrow
at = \sqrt{2gh} = \sqrt{2 E / m}
$$
were we used $s = s_0 + v_0 t + \frac{1}{2}at^2$. Now,
the impulse is defined as force times time
$$
J := F \cdot t = m a \cdot t
$$
where we again assumed that the force is constant during the time $t$. Thus, this yields 
$$
J = m \sqrt{2gh} = \sqrt{2 m E}
$$
Note, that if you do not use the approximation $s \ll h$ you end up with the impulse $J = m \sqrt{\frac{2gha}{g+a}}$. Again, by assuming $a\gg g$ you obtain the above solution. However, in principle the impulse depends on the acceleration $a$.
