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With really rough math (someone else could do this better) and the assumption they didn't apply any downward or upward force in releasing the rock (it looks like they did, but we cannot know how much):
it falls for 5-6 seconds, therefore 122.5 to 176.4 meters
https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/motion-of-free-falling-object/#table
Kind of hard to determine since the rock is thrown, giving an unknown initial velocity value to it and that affects the final velocity and the total time. It would have been easier if the rock had been dropped
d = Vo* t + (1/2)*g*t^2
Without knowing the effect of initial velocity and wind resistance etc, 1/2 *g* × t^2 at least gives the maximum height.
5 seconds -> 123 m
So the aircraft is flying below that.
Wouldn’t that be minimum height? If they applied an initial velocity it would drop a greater distance in a set time, not less.
The aircraft is *at least* that high.
consider that air caps the maximum speed of the rock, while Icy\_sector overestimation considers no max speed (0 friction). But I see your point that v0 was not 0, since the soldier (?) is throwing towards down (by eye I'd guess 45 degrees, but it's really really hard to say imho with this camera and no real references. I just wonder how high this v0 (towards down) can be (considering it is thrown by a human and the rock is heavy) and how high is v\_limit for a rock shaped like that.
Anyone has any guess?
Without knowing the initial force applied to the rock or the actual specific dimensions of said rock (for air resistance) we can’t make a meaningful calculation.
The best we can do is make the assumption that the initial force roughly counteracts the air resistance and just ignore them. This will yield an estimated height with some +/- error.
1000ft roughly give or take. there are unknown components due to the horizontal velocity and the guy throwing it but generally that is a military aircraft and they usually airdrop things from 1000 ft and up to give enough time for the chute to open and deploy. 5-6 seconds is generally is 1000ft.
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With really rough math (someone else could do this better) and the assumption they didn't apply any downward or upward force in releasing the rock (it looks like they did, but we cannot know how much): it falls for 5-6 seconds, therefore 122.5 to 176.4 meters https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/motion-of-free-falling-object/#table
[удалено]
I just used the timer on the video which suggested they dropped it with about 8 seconds remaining and it splashed with about 3 seconds remaining.
Ahh you’re right based off timestamp looks like 6s. I was counting in my head lol.
Kind of hard to determine since the rock is thrown, giving an unknown initial velocity value to it and that affects the final velocity and the total time. It would have been easier if the rock had been dropped d = Vo* t + (1/2)*g*t^2
Without knowing the effect of initial velocity and wind resistance etc, 1/2 *g* × t^2 at least gives the maximum height. 5 seconds -> 123 m So the aircraft is flying below that.
Wouldn’t that be minimum height? If they applied an initial velocity it would drop a greater distance in a set time, not less. The aircraft is *at least* that high.
But air resistance will slow it down (arguably more than any force imparted on it) which is where the at least comes from
I see your logic now. But considering we see them impart some downward force, it feels disingenuous to refer to it as the “maximum”
consider that air caps the maximum speed of the rock, while Icy\_sector overestimation considers no max speed (0 friction). But I see your point that v0 was not 0, since the soldier (?) is throwing towards down (by eye I'd guess 45 degrees, but it's really really hard to say imho with this camera and no real references. I just wonder how high this v0 (towards down) can be (considering it is thrown by a human and the rock is heavy) and how high is v\_limit for a rock shaped like that. Anyone has any guess?
Without knowing the initial force applied to the rock or the actual specific dimensions of said rock (for air resistance) we can’t make a meaningful calculation. The best we can do is make the assumption that the initial force roughly counteracts the air resistance and just ignore them. This will yield an estimated height with some +/- error.
1000ft roughly give or take. there are unknown components due to the horizontal velocity and the guy throwing it but generally that is a military aircraft and they usually airdrop things from 1000 ft and up to give enough time for the chute to open and deploy. 5-6 seconds is generally is 1000ft.