Maximum height of a particle launched at 2000 m/s? I have the radius, surface gravity, escape velocity, mass, and average density of a hypothetical planet. I also have the initial velocity of the object being launched. 
How do I determine the maximum height of an the object when the initial velocity is lower than the escape velocity?
 A: When you know the initial velocity, you know the kinetic energy - $E=\frac12 m v^2$. Now when projectiles travel just a short distance, you can say the the height $h$ is found by solving $mgh = \frac12 m v^2$ which gives you $h = \frac{v^2}{2g}$.
But when the height is sufficiently large (compared to the radius of the earth) you have to account for the fact that gravity gets less as you get higher. Instead of potential energy being proportional to $mgh$, we instead recognize that
$$V = -\frac{GMm}{r}$$
where $G$ is the universal gravitational constant, and $M$ is the mass of the earth, and $r$ the distance to the center of the earth ($r = R+h$). Because the gravitational acceleration at the surface of the earth is $g = \frac{GM}{R^2}$, we can write
$$\begin{align}
V &= -g\frac{mR^2}{r} \\
&= -g\frac{mR^2}{R+h}\\
&= -g \frac{mR}{1+\frac{h}{R}}\\
\end{align}$$
You can see that when $h<<R$ this reduces (by Taylor expansion of the denominator) to $V = -gR(1-\frac{h}{R} = -mgR + mgh$. Which is of course the familiar gravitational potential, plus a constant term (which we can ignore).
Now all you need to do is set the initial kinetic energy equal to the change in potential energy, using this equation without making the assumption that $h<<R$
Of course for your hypothetical planet of known density and radius, you would work with the planet's mass $M = \frac43 \pi R^3 \rho$ and use the universal gravitational constant $G=6.77\cdot 10^{-11} ~\rm{m^3 ~kg^{-1} ~s^{-1}}$. Also note that this answer ignores rotation of the planet, which is usually OK.
If all the above isn't enough, move your mouse over this spoiler:

Difference in potential energy $$\Delta V = GMm \left(\frac{1}{R}-\frac{1}{R+h}\right) = \frac12 m v^2$$. Solve for $h$.

A: 2km/s is very fast. If the planet has an Earth-like atmosphere there would be a significant amount of air resistance, proportional to either $v^2$ at high speed or $v$ at low speed, or something in between. The fact that the atmosphere gets thinner as you go further up makes it even more complicated.
There are also complications if the planet is rotating, or if the projectile is launched at an angle.
So it is best to assume that the planet has no atmosphere and is not rotating, and the projectile is launched straight up. Then maximum height is determined from the conservation of energy : the decrease in kinetic energy is equal to the increase in potential energy.
The decrease in KE is the initial KE of the projectile $\frac12mv^2$, since the final KE is zero. 
The PE at the surface of the planet and at maximum height are $-GMm/r_0$ and $-GMm/r_1$, where $r_0$ and $r_1$ are initial and final distances from the centre of the planet, maximum height is $h=r_1-r_0$, and $M$ is the mass of the planet. 
So
$\frac12mv^2 = GMm(\frac{1}{r_0}-\frac{1}{r_1})$
$v^2=2GM(\frac{1}{r_0}-\frac{1}{r_1})$
which you can solve to find $r_1$ then $h$. 
A: You will have to exploit work and energy concepts. You will have to use calculus because $F_g = G\frac{M_1M_2}{R^2}$. As you can see, the force varies with distance. Is this any help? Do you need more?
A: You seem to be given some redundant information as all you need is the escape speed and the radius of the planet.  
Using conservation of energy for the escape speed you can say that the kinetic energy plus the potential energy of the particle of mass at the surface of the planet is equal to the kinetic energy plus the potential energy of the particle at infinity.  
You can then set up an equivalent equation for the given velocity and so find the maximum height above the planet's surface.
