I don't think my post implied a touchdown speed of 5.6m/s, I used that as the initial velocity down to calculate the amount of fuel needed to land at under 1m/s.<br /><br /> But going on your more recent model with the liquids starting at 36m up and falling at 17m/s, I found the lander would need to accelerate up at about 4m/s/s to reduce it's touchdown speed to near zero. I added 3.7m/s/s to this, to account for the downward acceleration due to Mars's gravity, for a total of 7.7m/s/s acceleration those liquids need to apply to safely land. That might need a little bit more than 47kN thrust, as that applied to an 8000kg lander would give an acceleration of 5.875m/s/s or 5 & 7/8ths m/s/s.<br /><br /> With that acceleration, the initial velocity of 17m/s down, and the change in position of 36m, the liquid engines would need to provide a delta-velocity of 24m/s. To change the landers velocity by that much would require 192 000 newton-seconds thrust. I found the propellant mass by dividing that by 9.8 times the engine's isp, which I assumed to be 320s, and got 62kg of liquid propellant.<br /><br /> Actually I think your point about fuel consumption may be why we got different numbers. If 0.004528kg of propellant gives 1 newton-second of thrust, then dividing 1/0.004528 gives an exhaust velocity of 220.8m/s. This equates to a specific impulse of 22.5seconds. This is actually worse than the average cold gas thruster working on Nitrogen gas. With this specific impulse, the propellant mass would be closer to 870kg, so the change in acceleration per unit thrust due to loss of mass through the engines would become significant...<br /><br /> Other than the point about the liquid fuel required I can't find any major problems with what you've said. I don't think you've mentioned using a parachute in parallel, but that seems to be a non-issue from the decent speeds you've shown. And if the liquid engines are more efficient than the solids (might not be, dependin