On August 16, 1960, Joe Kittinger of the United States Air Force jumped from a h

Question

On August 16, 1960, Joe Kittinger of the United States Air Force jumped from a helium balloon from a height of 102,800 feet. After being in free fall for 4 minutes and 36 seconds, and falling for 85,300 feet, he opened his parachute and eventually landed safely on the ground. To analyze this situation, we will assume that Kittinger's initial velocity was zero, and that his acceleration was constant throughout the free fall. (This was most certainly not the case, but it gives us some idea about the motion.) Note the units we're looking for in the questions below.
(a) What was the magnitude of the acceleration during the free fall?
m/s2
(b) At the end of the free fall part of the motion, what was Kittinger's speed?
m/s


** the answer to A is NOT 9.81 and i dont know why

Explanation / Answer

85, 300 ft = 25999.44m 4min 36 sec = 276 sec a. H=.5at^2 25999.44 = .5(a)(276^2) a= .6826 m/s^2 b. Vf = Vi + at Vf = 0 + .6826(276) Vf = 188.401 m/s Hope this helps!

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