According to a newspaper account, a paratrooper survived a training jump from 1200 ft when his parachute failed to open but prov
ided some air resistance by ï¬apping unopen in the wind. allegedly he hit the ground at 100 mi/h after falling 8 seconds. test the accuracy of this account.
<span>Let's make a few assumptions.
1. The paratrooper's lowest speed will be just prior to impact.
2. Since the jump was from a relatively low altitude, the paratrooper used a static line and the parachute should have opened almost immediately upon jumping.
So let's convert 100 mi/h to ft/s
100 mi/h * 5280 ft/mi / 3600 s/h = 146.67 ft/sec
Given the 1st assumption above, the MAXIMUM distance the paratrooper would have fallen would be 8 seconds at 146.67 ft/s, so
8 s * 146.67 ft/s = 1173.36 ft
The calculated distance is close to the jump distance, which agrees with both assumptions 1 and 2. So this account does seem reasonable.
Additionally, looking for the speed of a parachutist doing a freefall in the belly-to-earth position with arms and legs outspread, they will generally reach a terminal velocity of 120 mi/h which is slightly faster than the 100 mi/h in the article. This too is in agreement with the defective parachute flapping and causing some extra air resistance.</span>
Explanation: Critical Thinking When a ball is thrown vertically up, it keeps going up until it hits a certain position, and then falls down. At the maximum point, the velocity is instantaneously zero.
Efficiency = StartFraction T Subscript h Baseline minus T Subscript C Baseline over T Subscript h Baseline EndFraction times 100. Efficiency equals T Subscript c Baseline minus T Subscript h Baseline over T Subscript h Baseline all times 100.
