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 63.1 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 83.8 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis): What are the x- and y-components of the velocity of the second skydiver, whose mass is 52.2 kg, immediately after separation? What is the change in kinetic energy of the system?

Explanation / Answer

Using conservation of momentum

83.8(5.93i + 5.25j) = 52.2( Vx i + Vy j)

Thus, Vx= 83.8*5.93/52.2 = 9.46 m/s

Vy = 83.8*5.25/52.2 = 8.43 m/s

There will be no change in the kinetic energy of the system, as there is no work done by an external force.

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