Part A) Two skydivers are holding on to each other while falling straight down a

Question

Part A) Two skydivers are holding on to each other while falling straight down at a common terminal speed of 51.10 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 89.30 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis):

Vix=4.430 m/s Viy=3.750 m/s Viz= 51.10 m/s

Part B) What are the x- and y-components of the velocity of the second skydiver, whose mass is 63.20 kg, immediately after separation?

Explanation / Answer

vix = 4.430m/s * 89.3/63.20 = 6.25 m/s
viy = 3.750m/s * 89.3/57.5 = 5.823 m/s
viz = 51.10 m/s

For second diver, KEi = ½ * 63.20kg * (51.10m/s)² = 82514 J
and KEf = ½ * 63.20kg * (51.10² + 6.25² + 5.823²)m²/s² = 84820 J
so the change is 2306 J

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