FYI, the relevant equation is to set the kinetic energy equal to the gravitational potential energy. This is written on a per-mass basis, because both kinetic energy and gravitational potential energy is proportional to mass.
$$ \frac{ G M m }{ r } = \frac{ 1 }{ 2} m v_{\infty}^2 $$
For Ceres:
$$ v_{\infty} = \sqrt{ \frac{ 2 G M }{r} } = \sqrt{ \frac{ 2 G ( 9.43 \times 10^{20} ) }{4.87 \times 10^5 m} } = 508.2 \frac{m}{s} $$
So obviously this is too high for someone to accomplish with human capabilities. It's also interesting to note that this follows a different mathematical form than the surface gravity. You could jump nearly 37 times as high on Ceres than on Earth, assuming the initial mechanics are fairly similar. This still isn't sufficient because Ceres' gravitational sphere of influence extends much further than this.
Also, the orbital velocity is lower by a factor of $\sqrt{2}$. So to obtain orbit you only need to reach $360 m/s$. While this is still not obtainable, you have a better shot at indefinitely departing from the ground if you exert force to the side, as opposed to straight up.
Also, note that Ceres has a relatively fast rotation. I calculate the equatorial velocity to be $93.7 m/s$, which would help you a good deal. If you get on the equator and jump in the direction of rotation, then you're down to $265 m/s$ to obtain orbit. Again, this is still unobtainable but it's the best shot you've got.
reduced-gravity-sports
tag, but interestingly all of your tags, evencrewed-spaceflight
actually apply nicely :) $\endgroup$