Relanding a Spaceship This problem is for my differentials class; I have a feeling it is much simpler than I am making it.
Here is the question: A spaceship is close to landing on a planet with constant gravity -g (-g because g is positive, but we decide positive direction is up). Ignore air resistance. The spaceship is currently (t=0) at height H and falling with velocity V. The retrorockets on the spaceship will provide constant acceleration a greater than g. We want to land the spaceship with velocity equal to zero when height is zero. What height should the retrorockets be activated?
Currently I have tried to split the problem into two parts: when the rocket is falling with gravity -g and when the retrorockets are providing acceleration a. I keep getting to a point where I make a separate time variable for the activation of the retrorockets, but I keep getting bogus answers. Any push in the right direction would be much appreciated!
 A: Without seeing your work, we can't see why you get an incorrect answer.  Yes, you should have two time intervals, one with rockets off and acceleration $-g$ and one with rockets on and acceleration $a$ (or $a-g$ as it is not clear if the acceleration of the rockets is net of gravity or not).  As a function of the time the rockets are off, you can compute the time, altitude, and downward velocity when the rockets come on.  Then as a function of the burn time, you can compute the altitude and velocity when the rockets turn off.  You get two equations in two unknowns, one from the altitude being zero and one from the velocity being zero.
A: Actually, I don't think any calculus is needed for the problem. 
Let's call the velocity of the spaceship when the rockets are activated $V_1$. Let's call the height (not displacement) of the spaceship (from the planet) when the rockets are activated $H_1$. 
Then, we know that $V_1^2=2H_1(a-g)$.
We also know that $V_1^2-V^2=2g(H-H_1)$. 
(This is due to the kinematics equation $V_f^2-V_i^2=2ax$. Note that, in this case, $x$ will be negative, since $x$ is displacement, not distance.)
Solving for $H_1$, we get that $H_1=\displaystyle\frac{V^2+2gH}{2a}$.
