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 61.90 m/ 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) 4.930 m/s 4250 m/s : 61.90 m/s What are the x- and y-components of the velocity of the second skydiver, whose mass is 63.20 kg, immediately after separation? Number V2,c=11-6.96 m/s Number What is the change in kinetic energy of the system? Number 292160 Joules Incorrect.

Explanation / Answer

I am solving the problem using conservation of momentum.

v2x = 4.93m/s * 89.3/63.20 = 6.97 m/s
v2y = 4.25m/s * 89.3/63.20 = 6.00 m/s
v2z = 61.90 m/s

For second diver, KEi = ½ * 63.20 kg * (61.9m/s)² = 121079 J

And, KEf = ½ * 63.20kg * (61.90² + 6.0² + 6.97²)m²/s² = 123752 J
Therefore, change in the kinetic energy of the system = 123752 - 121079 = 2673 J

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