Jumping on earth versus jumping on the moon Given the following problem:
On the moon the acceleration due to gravity is $g_m = 1.62 m/s^2$.  On earth, a person of mass $m = 80 kg$ manages to jump $1.4 m$.  Find the height this person will reach when jumping on the moon, if the person is wearing a spacesuit with mass $m = 124 kg$.
I am a little bit confused as to whether or not the given information regarding mass is actually needed here at all.  Assume that $v_0$ is equal in both jumps, and there is no rotational movement, can't we just use the formula for conservation of mechanical energy?
$$\frac{1}{2}m v_{f}^2 + mgh_f = \frac{1}{2}m v_{0}^2 + mgh_0$$
And here we can cancel the mass, $m$ since it appears in all terms.  So on earth we will have:
$$gh_f = \frac{1}{2} v_{0}^2$$
$$9.8 \cdot 1.4 = \frac{1}{2} v_{0}^2$$
$$v_0 = 5.2 m/s$$
Then on the moon, since we know $v_0$, we can then find $h_f$:
$$1.62 h_f = \frac{1}{2} \cdot (5.2)^2$$
$$h_f = 8.3 m$$
Would this not be an acceptable way to solve this?  If this is wrong, can anyone please explain why this is wrong conceptually? 
 A: Your solution is correct, if you assume that $v_0$ is the same on both moon and earth. The reason this seems counterintuitive is because, almost certainly, the jumper will not be able to generate the same $v_0$ on earth and moon. We can see this as followos.
The initial velocity is proportional to the integral of the force applied by the jumper's feet to the surface of the moon over time:
\begin{align}
\Delta v &= \frac{1}{m} \int_{t_i}^{t_f} dt \, F(t).
\end{align}
Assuming the force profile in time, $F(t)$ is the same on the earth and moon and is applied over the same time interval, $t_f-t_i$, we find that the initial velocity $v_0 = \Delta v$ is less on the moon owing to the larger total mass, $m_{jumper} + m_{suit}$.
A: Since this is homework, we are supposed to only give hints.  Your assumption that $v_0$ is the same is incorrect.  It is the only the force of the muscles that is the same.  Good luck!
A: Suppose we know the $g_e$ ($g$ at earth) and the man has to lift his
$c_m$ (center of mass) the same distance $s$ on the moon as on the
earth, until he loses contact with the ground.  Let $h_e$ and $h_m$ be
the heigths he reaches on earth and moon respectively.  Assuming the
force is the same until he loses contact to the ground, on the moon
and on earth then the work $W$ that produces during "take off" is
also the same.  This work is used to bring him at maximum height.
Then on Earth: $W=m g_e h_e$,
on Moon: $W = (m + m_s) g_m h_m$,
and from above, $m g_e h_e=(m+m_s)g_m
h_m$ and 
$h_m = (m / (m+m_s)) (g_e/g_m) h_e$.
