Two skydivers are holding on to each other while falling straight down at a comm

Question

Two skydivers are holding on to each other while falling straight down at a common terminal speed of 61.9 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 94.8 kg) has the following velocity components (with ''straight down'' corresponding to the positive z-axis): vl, x = 5.43 m / s vl,y. = 4.25 m / s v1,z = 61.9 m/s What are the x- and y-components of the velocity of the second skydiver, whose mass is 63.2 kg, immediately after separation? V2,x = V2,y= What is the change in kinetic energy of the system?

Explanation / Answer

conservation of momentum in x direcction

m1v1x=m2v2x

94.8*5.43=63.2*v2x

v2x=8.145m/s

conservation of momentum in y direcction

m1v1y=m2v2y

94.8*4.25=63.2*v2y

v2y=6.375 m/s

Change in kinetic energy=5634.366 J

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