2 Pointed users to uhoh's graphics edited Aug 20 '17 at 16:38 Quietghost 1,52077 silver badges2121 bronze badges Your intuition about the combination of the two is leading you astray. In fact, the speed is a combination of the two. 1600km/h is 450m/s, which is practically stationary in terms of orbital mechanics. Also, remember that any graphics you see are not to scale. For the more detailed answer: The boostback burn has to negate all the horizontal velocity and add reverse horizontal velocity to come back. Given that it has probably flown downrange on the order of 50km by stage separation, it needs to supply enough horizonal velocity to come back all of those 50km. It can be done quickly or slowly. If it is done quickly, a lot of fuel will be used in adding horizontal reverse velocity and then cancelling it out again. If it is done slowly, then the rocket might fall back to earth before it returns close enough to the pad. But the magic of the boostback burn is that it preserves vertical upwards velocity, and so a large amount of time is gained for the horizontal trip to the pad. But this time gained is on the order of a minute or two, and so the horizontal velocity reflects that. Just some rough numbers: $$50\mathrm{km} / 120\mathrm{sec} \approx 420\mathrm{m/s} \approx 1500 \mathrm{km/h}$$ So the speeds shown in the webcast at highest point are in fact spot on. Edit: uhoh has put together some awesome graphics, showing just how aggressive the horizontal trajectory is. Check uhoh's answer out! Your intuition about the combination of the two is leading you astray. In fact, the speed is a combination of the two. 1600km/h is 450m/s, which is practically stationary in terms of orbital mechanics. Also, remember that any graphics you see are not to scale. For the more detailed answer: The boostback burn has to negate all the horizontal velocity and add reverse horizontal velocity to come back. Given that it has probably flown downrange on the order of 50km by stage separation, it needs to supply enough horizonal velocity to come back all of those 50km. It can be done quickly or slowly. If it is done quickly, a lot of fuel will be used in adding horizontal reverse velocity and then cancelling it out again. If it is done slowly, then the rocket might fall back to earth before it returns close enough to the pad. But the magic of the boostback burn is that it preserves vertical upwards velocity, and so a large amount of time is gained for the horizontal trip to the pad. But this time gained is on the order of a minute or two, and so the horizontal velocity reflects that. Just some rough numbers: $$50\mathrm{km} / 120\mathrm{sec} \approx 420\mathrm{m/s} \approx 1500 \mathrm{km/h}$$ So the speeds shown in the webcast at highest point are in fact spot on. Your intuition about the combination of the two is leading you astray. In fact, the speed is a combination of the two. 1600km/h is 450m/s, which is practically stationary in terms of orbital mechanics. Also, remember that any graphics you see are not to scale. For the more detailed answer: The boostback burn has to negate all the horizontal velocity and add reverse horizontal velocity to come back. Given that it has probably flown downrange on the order of 50km by stage separation, it needs to supply enough horizonal velocity to come back all of those 50km. It can be done quickly or slowly. If it is done quickly, a lot of fuel will be used in adding horizontal reverse velocity and then cancelling it out again. If it is done slowly, then the rocket might fall back to earth before it returns close enough to the pad. But the magic of the boostback burn is that it preserves vertical upwards velocity, and so a large amount of time is gained for the horizontal trip to the pad. But this time gained is on the order of a minute or two, and so the horizontal velocity reflects that. Just some rough numbers: $$50\mathrm{km} / 120\mathrm{sec} \approx 420\mathrm{m/s} \approx 1500 \mathrm{km/h}$$ So the speeds shown in the webcast at highest point are in fact spot on. Edit: uhoh has put together some awesome graphics, showing just how aggressive the horizontal trajectory is. Check uhoh's answer out! 1 answered Aug 16 '17 at 15:58 Quietghost 1,52077 silver badges2121 bronze badges Your intuition about the combination of the two is leading you astray. In fact, the speed is a combination of the two. 1600km/h is 450m/s, which is practically stationary in terms of orbital mechanics. Also, remember that any graphics you see are not to scale. For the more detailed answer: The boostback burn has to negate all the horizontal velocity and add reverse horizontal velocity to come back. Given that it has probably flown downrange on the order of 50km by stage separation, it needs to supply enough horizonal velocity to come back all of those 50km. It can be done quickly or slowly. If it is done quickly, a lot of fuel will be used in adding horizontal reverse velocity and then cancelling it out again. If it is done slowly, then the rocket might fall back to earth before it returns close enough to the pad. But the magic of the boostback burn is that it preserves vertical upwards velocity, and so a large amount of time is gained for the horizontal trip to the pad. But this time gained is on the order of a minute or two, and so the horizontal velocity reflects that. Just some rough numbers: $$50\mathrm{km} / 120\mathrm{sec} \approx 420\mathrm{m/s} \approx 1500 \mathrm{km/h}$$ So the speeds shown in the webcast at highest point are in fact spot on.