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 54.70 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 94.80kg) has the following velocity components (with "straight down" corresponding to the positive z-axis): V_1,x = 4.430 m/s V_1,y 4.250 m/s V_1,z = 54.70 m/s What are the x- and y-components of the velocity of the second skydiver, whose mass is 52.20 kg, immediately after separation? What is the change in kinetic energy of the system?

Explanation / Answer

Conservation of momentum in x direction,

94.8*4.43 + 52.2*v2,x = 0

v2,x = -8.045 m/s

Conservation of momentum in y direction,

94.8*4.25 + 52.2*v2,y = 0

v2,y = -7.718 m/s

Change in kinetic energy of first skydiver = (1/2)m(v21,x + v21,y) = (1/2)*94.8*(4.432 + 4.252) = 1786.38 J

Change in kinetic energy of second skydiver = (1/2)m(v22,x + v22,y) = (1/2)*52.2*(8.0452 + 7.7182) = 3243.96 J

Total change in kinetic energy = 1786.38 + 3243.96 = 5030.34 J

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