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 55.9 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 89.3 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis): V1,x = 7.43 m/s V1,y = 5.75 m/s V1,z = 55.9 m/s What are the x- and y- components of the velocity of the second skydiver, whose mass is 57.7 kg, immediately after separation? V2,x = ? m/s V2,y =? m/s What is the change in kinetic engery of the system? in Joules

Explanation / Answer

Momentum will be conserved

Initially total momentum = (89.3 + 57.7 ) * 55.9 = 8217.3 kgm/s in positive z direction

Now m1V1x + m2V2x = 0 => V2x = - 89.3*7.43/57.7 = -11.499 m/s

m1V1y + m2V2y = 0 => V2y = -89.3*5.75/57.7 = -8.899 m/s

Initial kinetic energy = (89.3 +57.7 )* 55.9^2/2 = 229673.535 J

Final kinetic energy = 89.3 * ( 7.43^2 + 5.75^2 +55.9^2)/2 + 55.7 * ( 11.499^2 + 8.899^2 + 55.9^2 )/2 = 479428.237 J

Change in KE = 479428.237 - 229673.535 = 249754.702 J

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