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 53.5 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): v1,x =7.93 m/s v1,2 =4.25 m/s v1,z,.= 53.5 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? What is the change in kinetic energy of the system?

Explanation / Answer

conservation of momentum in the x:
0=94.8*7.94 + 63.2*v
vx=-11.91 m/s

y:
0 = 94.8*4.25+63.2*v
vy=-6.375 m/s

z: 53.5 m/s

dKE = KEf - KEi = 1/2*94.8*(7.93^2+4.25^2+53.5^2) + 1/2*63.2*(11.91^2+6.375^2+53.5^2) -1/2*94.8*53.5^2 -1/2*63.2*53.5^2=9604 J

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