A friend posted a story from the internet that goes:
A couple decided to commit suicide. They'd had some rough times and chose to jump from a building. When they got to the top, they counted to 3 and the woman jumps. The man watches her fall for about 8 seconds when a parachute opens. Who betrayed whom?
Her comment? "Must be a tall building!"
Of course I took it upon myself to figure out just how tall that building is. This is more interesting than a basic equations of motion problem because the use of a parachute necessitates allowing for variable forces, both during the initial freefall and then during the slowing down process after the chute opens. To frame up the initial fall, we can write, generally:
[1] d/dt D(t) = v(t)
[2] d/dt v(t) = a(t)
[3] a(t) = F(t)/m
These are our base, non-specific equations. The force diagram on the woman has 2 components: mg down (which we'll define as the positive direction), and kv up, where k is some constant aerodynamic coefficient; air resistance can be crudely modeled as proportional to velocity. We have additional data as well. We know that a person in freefall will reach a terminal velocity, Tv, and that value is generally stated as being about 50m/s. I'll use this value for the calculations, but if you feel they aren't right, you're welcome to use alternate values in the final equations. So, at the point of terminal velocity we know that kTv = mg, and thus k = mg/Tv.
[3.1] a(t) = (mg-kv(t))/m
[3.2] a(t) = (mg-mgv(t)/Tv)/m
[3.3] a(t) = g - gv(t)/Tv
We can do a quick sanity check here: as v(t) -> Tv, acceleration indeed goes to zero as expected. The next step involves solving the differential equations [2] and [1]. While this could be a fun exercise in math process, I'll defer to Wolfram Alpha diff eq solver. Known boundary conditions (notably that velocity and distance are both zero at time=0) will be used to solve for constants that appear as a result of integration steps.
[2.1] d/dt v(t) = g - gv(t)/Tv
[2.solved] v(t) = c1 exp(-gt/Tv) + Tv
[2.boundary] v(0) = 0 = c1 + Tv, thus c1 = -Tv
[2.final] v(t) = Tv - Tv exp(-gt/Tv)
[1.1] d/dt D(t) = Tv - Tv exp(-gt/Tv)
[1.solved] D(t) = c2 + Tv t + Tv Tv exp(-gt/Tv) / g
[1.boundary] D(0) = 0 = c2 + Tv Tv / g, thus c2 = -Tv Tv / g
[1.final] D(t) = Tv Tv exp(-gt/Tv) / g + Tv t - Tv Tv /g
Solving for distance fallen and speed after 8 seconds yields her status as:
D(8) = 196 meters down
v(8) = 40 m/s
We're now at the stage where her parachute is opening. Based on a combination of movie scenes and the shortest base jump I could find, it appears it takes about 2 seconds for the chute to open. During this time I'll assume it has negligible braking force, and I'll simplify and assume the woman would continue to fall at v(8) = 40 m/s and thus finish this stage at 276 meters down and still going 40 m/s. Clearly I've simplified here, but due to the short duration of the stage, the effects are not terribly material on the final calculations. Others suggest the time could be as low as 1.2 seconds (derived by dividing the claimed 200 feet distance to open by terminal velocity of 50 m/s). That would put a lower-bound on her fall at 244 meters.
The final stage is to figure out how long it takes for the chute drag to slow her down to a speed she'll survive. We can reuse equations [1.solved] and [2.solved] with new boundary conditions. Tv in this case is the descent speed with the chute open, which I'll call about 5 m/s.
[2.boundary.stage3] v(0) = 40 = c1 exp(-gt/Tv) + Tv = c1 + Tv, thus c1 = 35
[2.final.stage3] v(t) = 35 exp(-gt/Tv) + Tv
[1.boundary.stage3] D(0) = 0 ** = c2 + Tv t - 35 Tv exp(-gt/Tv) / g = c2 - 35 Tv / g, thus c2 = 35 Tv / g
** - I'm re-referencing her position here as zero, this can be trivially added/shifted to the previous stages.
[1.final.stage3] D(t) = 35 Tv / g - 35 Tv exp(-gt/Tv) / g + Tv t
The last thing to figure out is how long it takes her to slow down to a speed she can still land reasonably safely. I'm not sure what the number is, but let's say 2 Tv = 10 m/s. That's the equivalent of jumping from about 15 feet. It'd hurt, but she'd survive. Using [2.final.stage3] and solving for t:
[2.final.stage3.survival1] v(t) = 2 Tv = 35 exp(-gt/Tv) + Tv
[2.final.stage3.survival1.1] Tv = 35 exp(-gt/Tv)
[2.final.stage3.survival1.2] Tv ln (Tv / 35) / -g = t
[2.final.stage3.survival1.3] t ~= 1 second
As an alternative, suppose we want to get to 1.2 Tv for a practically terminal velocity landing
[2.final.stage3.survival2] v(t) = 1.2 Tv = 35 exp(-gt/Tv) + Tv
[2.final.stage3.survival1.1] Tv = 175 exp(-gt/Tv)
[2.final.stage3.survival1.2] Tv ln (Tv / 175) / -g = t
[2.final.stage3.survival1.3] t ~= 1.8 seconds
Plugging in our two values will give us a reasonable upper and lower bound for how much distance she needs after the chute opens.
[1.final.stage3.survival1] D(1) = 20.3 m
[1.final.stage3.survival2] D(1.8) = 26.3 m
The final stage is quite rapid, and slowing down considerably more only costs an extra 6 meters. Let's assume for comfort, she'd want to get to the slower speed. Adding up the 3 stages yields 196 in freefall, 48 to 80 while the chute is opening, and another 26 to slow down. To be on the safe-ish side, let's take the max of each stage and say the building needs to be at least 196 + 80 + 26 = 302 meters (991 feet) tall. In essence, this could be done off, with rounding, any building over 1000 feet (or nominally 100 stories) tall.
There are only a few buildings in America that this story could plausibly have happened on. There are only 17 that are formally tall enough, and a few of those are coming within feet of the exact minimum height. However, a number of those don't qualify because they don't have a platform that's high enough. For example, it's doubtful that anyone could jump outwards enough from high enough on the Empire State Building or Bank of America Tower to avoid significantly outcropping lower layers. The New York Times and Chrysler buildings definitely fall into this category, and judging from other photos, the Bank of America Plaza and US Bank Tower are similarly problematic. The bottom end of the tall-enough buildings all seem to be box-shaped enough that they would support our scenario. All in all, this could happen in maybe 10 or 11 specific places, but would have to be in New York, Chicago or Houston.
What's fun to note is that even from full terminal velocity, the distance needed to open a chute and slow down to safe landing speed is under 150 meters. Scaling this up to the tallest building in the world, Using [1.final] and D(t) = 680 allows for 18 seconds of free-fall! Intense!
A couple decided to commit suicide. They'd had some rough times and chose to jump from a building. When they got to the top, they counted to 3 and the woman jumps. The man watches her fall for about 8 seconds when a parachute opens. Who betrayed whom?
Her comment? "Must be a tall building!"
Of course I took it upon myself to figure out just how tall that building is. This is more interesting than a basic equations of motion problem because the use of a parachute necessitates allowing for variable forces, both during the initial freefall and then during the slowing down process after the chute opens. To frame up the initial fall, we can write, generally:
[1] d/dt D(t) = v(t)
[2] d/dt v(t) = a(t)
[3] a(t) = F(t)/m
These are our base, non-specific equations. The force diagram on the woman has 2 components: mg down (which we'll define as the positive direction), and kv up, where k is some constant aerodynamic coefficient; air resistance can be crudely modeled as proportional to velocity. We have additional data as well. We know that a person in freefall will reach a terminal velocity, Tv, and that value is generally stated as being about 50m/s. I'll use this value for the calculations, but if you feel they aren't right, you're welcome to use alternate values in the final equations. So, at the point of terminal velocity we know that kTv = mg, and thus k = mg/Tv.
[3.1] a(t) = (mg-kv(t))/m
[3.2] a(t) = (mg-mgv(t)/Tv)/m
[3.3] a(t) = g - gv(t)/Tv
We can do a quick sanity check here: as v(t) -> Tv, acceleration indeed goes to zero as expected. The next step involves solving the differential equations [2] and [1]. While this could be a fun exercise in math process, I'll defer to Wolfram Alpha diff eq solver. Known boundary conditions (notably that velocity and distance are both zero at time=0) will be used to solve for constants that appear as a result of integration steps.
[2.1] d/dt v(t) = g - gv(t)/Tv
[2.solved] v(t) = c1 exp(-gt/Tv) + Tv
[2.boundary] v(0) = 0 = c1 + Tv, thus c1 = -Tv
[2.final] v(t) = Tv - Tv exp(-gt/Tv)
[1.1] d/dt D(t) = Tv - Tv exp(-gt/Tv)
[1.solved] D(t) = c2 + Tv t + Tv Tv exp(-gt/Tv) / g
[1.boundary] D(0) = 0 = c2 + Tv Tv / g, thus c2 = -Tv Tv / g
[1.final] D(t) = Tv Tv exp(-gt/Tv) / g + Tv t - Tv Tv /g
Solving for distance fallen and speed after 8 seconds yields her status as:
D(8) = 196 meters down
v(8) = 40 m/s
We're now at the stage where her parachute is opening. Based on a combination of movie scenes and the shortest base jump I could find, it appears it takes about 2 seconds for the chute to open. During this time I'll assume it has negligible braking force, and I'll simplify and assume the woman would continue to fall at v(8) = 40 m/s and thus finish this stage at 276 meters down and still going 40 m/s. Clearly I've simplified here, but due to the short duration of the stage, the effects are not terribly material on the final calculations. Others suggest the time could be as low as 1.2 seconds (derived by dividing the claimed 200 feet distance to open by terminal velocity of 50 m/s). That would put a lower-bound on her fall at 244 meters.
The final stage is to figure out how long it takes for the chute drag to slow her down to a speed she'll survive. We can reuse equations [1.solved] and [2.solved] with new boundary conditions. Tv in this case is the descent speed with the chute open, which I'll call about 5 m/s.
[2.boundary.stage3] v(0) = 40 = c1 exp(-gt/Tv) + Tv = c1 + Tv, thus c1 = 35
[2.final.stage3] v(t) = 35 exp(-gt/Tv) + Tv
[1.boundary.stage3] D(0) = 0 ** = c2 + Tv t - 35 Tv exp(-gt/Tv) / g = c2 - 35 Tv / g, thus c2 = 35 Tv / g
** - I'm re-referencing her position here as zero, this can be trivially added/shifted to the previous stages.
[1.final.stage3] D(t) = 35 Tv / g - 35 Tv exp(-gt/Tv) / g + Tv t
The last thing to figure out is how long it takes her to slow down to a speed she can still land reasonably safely. I'm not sure what the number is, but let's say 2 Tv = 10 m/s. That's the equivalent of jumping from about 15 feet. It'd hurt, but she'd survive. Using [2.final.stage3] and solving for t:
[2.final.stage3.survival1] v(t) = 2 Tv = 35 exp(-gt/Tv) + Tv
[2.final.stage3.survival1.1] Tv = 35 exp(-gt/Tv)
[2.final.stage3.survival1.2] Tv ln (Tv / 35) / -g = t
[2.final.stage3.survival1.3] t ~= 1 second
As an alternative, suppose we want to get to 1.2 Tv for a practically terminal velocity landing
[2.final.stage3.survival2] v(t) = 1.2 Tv = 35 exp(-gt/Tv) + Tv
[2.final.stage3.survival1.1] Tv = 175 exp(-gt/Tv)
[2.final.stage3.survival1.2] Tv ln (Tv / 175) / -g = t
[2.final.stage3.survival1.3] t ~= 1.8 seconds
Plugging in our two values will give us a reasonable upper and lower bound for how much distance she needs after the chute opens.
[1.final.stage3.survival1] D(1) = 20.3 m
[1.final.stage3.survival2] D(1.8) = 26.3 m
The final stage is quite rapid, and slowing down considerably more only costs an extra 6 meters. Let's assume for comfort, she'd want to get to the slower speed. Adding up the 3 stages yields 196 in freefall, 48 to 80 while the chute is opening, and another 26 to slow down. To be on the safe-ish side, let's take the max of each stage and say the building needs to be at least 196 + 80 + 26 = 302 meters (991 feet) tall. In essence, this could be done off, with rounding, any building over 1000 feet (or nominally 100 stories) tall.
There are only a few buildings in America that this story could plausibly have happened on. There are only 17 that are formally tall enough, and a few of those are coming within feet of the exact minimum height. However, a number of those don't qualify because they don't have a platform that's high enough. For example, it's doubtful that anyone could jump outwards enough from high enough on the Empire State Building or Bank of America Tower to avoid significantly outcropping lower layers. The New York Times and Chrysler buildings definitely fall into this category, and judging from other photos, the Bank of America Plaza and US Bank Tower are similarly problematic. The bottom end of the tall-enough buildings all seem to be box-shaped enough that they would support our scenario. All in all, this could happen in maybe 10 or 11 specific places, but would have to be in New York, Chicago or Houston.
What's fun to note is that even from full terminal velocity, the distance needed to open a chute and slow down to safe landing speed is under 150 meters. Scaling this up to the tallest building in the world, Using [1.final] and D(t) = 680 allows for 18 seconds of free-fall! Intense!
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