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 52.30 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 94.80 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 52.20 kg, immediately after separation? What is the change in kinetic energy of the system?

Explanation / Answer

When they push away from each other

X component of Velocity -

Mx = m1 * Vx - m2 * Vx2 = 0
94.80 * 4.930 = 52.2 * Vx2
Vx2 = 8.95 m/s

Y component of Velocity -
My = m1 * Vy - m2 * Vy2 = 0
94.80 * 4.750 = 52.2 * Vy2
Vy2 = 8.63 m/s

Vz2 = 52.30 m/s

Change in K.E = K.E Final - K.E initial

K.Ein = 1/2 * (94.8 + 52.2) * 52.30^2
K.Efi = 1/2 * 94.8 * (4.930^2 + 4.750^2 + 52.30^2) + 1/2 * 52.2 * (8.95^2 + 8.63^2 + 52.30^2)
Change in K.E = K.Efin - K.Ein
Change in K.E = 6256.0 J

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