Understanding rocket problem intuitively 
A rocket is trying to land on a planet. The mass of the rocket is $1\,\rm kg$, and the gravitational acceleration of the planet is $1\,\rm m/s^2$. The rocket starts the free fall at $20\,\rm m$ above the surface of the planet (initial velocity is $0$), and can use the thrust for $2\,\rm s$ (the force of thrust is $1\,\rm N$).
When is the most reasonable height at which the rocket uses its thrust for two seconds? (By the way, we ignore the loss of mass due to the use of the thrust.)

I solved the question, but I'm not satisfied. I don't quite understand it intuitively.
Someone said $W = Fs$, and since $F$ (thrust) is the same, when $s$ (distance moved) is the greatest, the work done by the thrust, to counteract gravity, would be the greatest. Therefore, the most reasonable height to start using the thrust is when the height at which the rocket would end using its thrust is when it reaches the ground (the calculation to find the actual value of the height is very complicated, so I'll skip (it's not the main point of my question).
When I first tried to solve this, I thought that the chemical energy of the thrust would be used to counteract gravity, and since the chemical energy of the the thrust does not change by the velocity at which it moves, I thought that the height at which the thrust is used does not matter, as the total energy (potential energy due to the gravity + the kinetic energy of the rocket - the chemical energy of the thrust) stays the same, the final velocity would be the same, but this is not the answer.
Can anyone please help me why I may be wrong?
 A: With this sort of problem it is often worth sketching a velocity against time graph.

The constraints are that the rocket has to travel $20$ m, will accelerate at $1$  ms$^{-2}$ and with the thruster on the rocket will descend at constant velocity.
So the trick is to get the rocket to fall as far as possible when its thruster is on as is shown in the diagram below.  
So the equation to solve is $\frac 1 2 t^2 +2t =20 \Rightarrow t \approx 4.63$ s
If you put that constant velocity rectangle anywhere else the distance travelled at constant velocity will be less and so the distance travelled accelerating will be more.
So more time will be spent accelerating and so the landing velocity will be larger.
A: It is possible to use law of conservation of energy to understand what is going on here. But simple arguments such as "chemical energy opposes gravity" would not help.
So the fuel was burned, it's chemical energy is released. The amount of energy released is the same no matter if fuel was burned in the beginning or in the end of the flight. Where did this energy go?


*

*the gas thrown by rocket is heated up: $E_1$

*kinetic energy of the gas has changed: $E_2$

*kinetic energy of the rocket has changed: $E_3$
We want to make $E_3$ as small as possible (making it negative would be nice: that would mean that kinetic energy of the rocket has decreased).
$E_1$ does not depend on when the fuel was burned.
So the only way to decrease $E_3$ is to increase $E_2$ as much as possible. That is to increase the kinetic energy of outgoing gas as much as possible. Does the kinetic energy of gas depend on anything? It does.
The rocket engines throw gas with some velocity $V$. If the velocity of the rocket is $v$ than the change of gases kinetic energy would be:
$$m\frac{(v+V)^2}{2} - m\frac{v^2}{2} = \frac{m}{2} (2Vv + V^2)$$
See? The greater the speed $v$ of the rocket the more kinetic energy gains the gas when rocket's engines throw it forward with additional speed $V$!
So if the rocket burns the fuel in the end of the flight when it's speed is at maximum the kinetic energy of gas is maximized. And the kinetic energy of the rocket is minimized.
A: As far as I see this, the thrusters provide a thrust equal to the weight of the rocket, so firing them can only make the nett force on the rocket zero for $2$ seconds, which means the rocket will not accelerate for $2$ seconds. So, the thrusters should be fired at $2$ seconds to land.
According to my calculations the height should be $9.26\,\rm m$ (correct me if I am wrong). Also note that the velocity at this point will be $4.63\,\rm m/s$. The impact velocity can not be zero in any case. The impact velocity could be zero if the thrust provided was greater than the weight of the rocket.
Also, in a realistic scenario, the impact velocity might be zero as the weight of the rocket would reduce with the fuel being used up.
A: To minimise the impact speed, there are two ways to think about this:


*

*You want to maximise the work done on the rocket:
$$\mathrm{Work = Force \times Distance}$$

*You need to minimise the time spent falling.
A: You're right that the total energy will always be conserved.  But in this type of problem, you're not given a fixed amount of energy to use, instead you're given a fixed amount of impulse (force multiplied by time).  In effect, the efficiency (how much change in energy it produces) of the rocket is changing, depending on when it is used.  If the rocket is used when the ship is traveling quickly ($v$ is large), then it works more efficiently to slow the ship down.
A: Assuming the goal is to minimise impact speed then you should time your burn so that it ends when you land.
Why? because all the time you are in flight you are gaining speed due to gravity. Burning earlier makes your flight longer and thus increases your impact speed.
