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[TimeSpaceLightForce]

Five balloonist will travel up to a 1000 ft. high plateau to see the sunrise.
Each carry a rope and a 1, 2, 3, 4, 5 Lbs sandbag so that they ascend
at 5, 4, 3, 2, 1 ft/sec respectively.(the hellium balloons are similar)
In order to arrive altogether at the same time to the top they may use the
ropes to pass or exchange their sandbags with the others at any time.
Starting all on the ground, how would they plan the trip with minimum
passes or exchanges?

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[austinm]

after 166 2/3 sec balloons 1&5 and 2&4 should swap sandbags.

(Altitudes at this time will be 166'8", 333'4", 500', 666'8", 833'4", requiring a 500' pass between balloons 1&5.)

After another 166 2/3 sec all five balloons will have reached 1000': only two swaps or four bag-moves-positions, for a total ascent time of 333 1/3 sec (as in Bonanova's derivation).

Why not just suspend all five sandbags centrally, tethered spiderweb-style to all of the balloons. Assuming the tethers come up from the central load at 30 deg off-vertical the balloons--at the vertices of a regular pentagon with radius 250'--would enjoy nearest-neighbor separation of ~294'.

Good enough? (I don't know--only been ballooning once.) But it requires no swaps, and for 433' of the journey the balloons are under no load.

Presuming (from the OP's linear progression) that the unencumbered ascent rate will be 6fps, that shaves ~94.5 sec off the journey, for an ascent time of ~239sec!

Lets take two of the 500' ropes, tie end-to-end (magically using no length doing so), tether all five sandbags to one balloon, and the whole fleet can ascend at 6fps until they all reach 1000' ~144 sec. into the trip. No swaps!

[bonanova]

Brute force solution has each balloon carrying each weight for the same length of time.

Thus at four equals time intervals (dt) the weights are passed cyclically among the balloons.

The sum of the distances travelled in these time intervals must be 1000 feet.

Thus, Sum (d_i) = Sum (r_i) dt = 1000; dt = 1000/Sum (r_i) = 1000/15 = ~66.7 seconds.

The balloons all arrive at the plateau in 5 dt = 1000/3 = ~333.3 seconds.

This is not necessarily optimal.

[DeGe]

2 exchanges needed.

The balloon with bag number 3 is not involved in any exchange. The ballons with 5 and 1 lb bags exchange once and the ballons with 4 and 2 lbs bags exchange once.
Total time taken is 1000/3 = 333.3 secs; Both 4/2 and5/1 exchange the bags at exactly 166.67 secs.

[TimeSpaceLightForce]

All right!