Kittinger entered record books when he stepped from the gondola of a helium floa

Question

Kittinger entered record books when he stepped from the gondola of a helium floating an altitude of 31,330m. The air is so thin at this altitude that it would make for a moderate laboratory vacuum on the surface of the earth. With little atmosphere, the sky is essentially black and the sun's radiation is unusually intense despite polar temperatures. The density of air at 30 km is roughly 1.5 percent that at sea level and thus air drag is essentially negligible. According to kittinger's report in national geographic, he was in free fall from 102,800 to 96,000 ft and then experienced no noticeable change in acceleration for an additional 6,000 ft despite having deployed his stabilization parachute. This gave him an unprecedented 12,800 feet over which accelerate. At such extreme altitudes the acceleration due to gravity is not the standard 9.8 m/s^2, but the slightly lower value of 9.72 m/s^2. Let us assume he freefell 12,800 ft with acceleration of magnitude 9.72 m/s^2 and then when his main parachute opened his overall acceleration instantly dropped to zero. 1)What is the distance he essentially freefell, in meters? 2)How long did it take him to fall this distance, in seconds? 3)what was his speed after falling this distance, in m/s?,in mph? 4)how long to fall the remaining distance to earth assuming the speed in question 4? 5what is the overall acceleration being assumed in question 5? 6with what speed does he hit the ground

Explanation / Answer

Well part 1 is easy enough, it wants you to convert 12,800 feet to metres. 12800 * 12 * 2.54 / 100 = 3,901.44 m 3,900 m to 3 significant figures. Now it's mostly constant acceleration formula, for question 2 you can use the forumla: s = ut + 0.5*a*t^2 Where: s = displacement or distance (3,900 m) u = initial speed (0 m/s) v = final speed (unknown as yet) a = acceleration (9.72 m/s^2) t = time (unknown as yet) For question 3 you need another such formula: v^2 = u^2 + 2as Then convert to mph, I'll leave you to figure that out. For question 4 it's a standard speed = distance / time For question 5 the answer is given in the initial question info. For question 6 the answer is the same as for question 3 as you are assuming he stays at that speed all the way down.

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