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 58.30 m/s. Suddenly, they push away from each other. Immediately after separation, the first skydiver (who has a mass of 83.80 kg) has the following velocity components (with "straight down" corresponding to the positive z-axis): v_1, x = 4.430 m/s v_1, y = 3.750 m/s v_1, z = 58.30 m/s What are the and y-components of the velocity of the second skydiver, whose mass is 63.20 kg, immediately after separation? v_2, x = v_2, y = What is the change in kinetic energy of the system?

Explanation / Answer

given data:
terminal speed = 58.30 m/s----------------54.70
mass of 1st skydiver = 83.80 kg----------------94.80
v1x =4.430 m/s--4.930
v1y = 3.750 m/s----4.250
v1z = 58.30 m/s------------54.70
mass of 2nd skydiver = 63.20 kg---57.70

There are no net external forces applied (system acceleration along all three axes is zero), so conservation of momentum applies in all three directions.
x:
(83.80 + 63.20)* 0 = 83.80 * 4.430 + 63.20 * V2x
V2x = - 5.87 m/s
y:
(83.80 + 63.20) * 0 = 83.80 * 3.750 + 63.20 * V2y
V2y = - 4.972 m/s
z:
(83.80 + 63.20) * 58.30 = 83.80 * 58.30+ 63.20* V2z
V2z = 58.3 m/s
initial KE = (1/2)* (83.80+63.20)* (58.30)^2 = 249818 J

final KE1 = (1/2) * 83.80 * (4.430^2 + 3.750^2 + 58.30^2)= 143825 J
final KE2 = (1/2)* 63.20 * (5.87^2+ 4.972^2 + 58.3^2) = 109275 J
for a total final KE = 253100
meaning that the increase in KE is 34550 J
This is the work that they do in pushing off from each other.

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