Two skydivers are holding on to each other while falling straight down at a common terminal speed of 57.10 m / s. Suddenly, they push away from each other. Immediately after separation, the first skydiver, who has a mass m 1 of 94.80 kg, has a velocity v 1 with the following components ("straight down" corresponds to the positive z - axis). v 1 , x = 5.430 m / s v 1 , y = 4.250 m / s v 1 , z = 57.10 m / s What are the x - and y - components of the velocity v 2 of the second skydiver, whose mass m 2 is 57.70 kg, immediately after separation? Assume that there are no horizontal forces of air resistance acting on the skydivers.

-8.88 m/s on x and -6.98 m/s on y

Explanation:

Given

Common terminal velocity, v(t) = 57.1 m/s

Mass of the first skydiver, m1 = 94.8 kg

Velocity of the skydiver

v1(x) = 5.43 m/s

v1(y) = 4.25 m/s

v1(z) = 57.1 m/s

Mass of the second skydiver, m2 = 57.7 kg

Now, in order to find the components of velocity along X and Y axis:

Before they got separated, the momentum along X - axis is zero.

After they got separated, the momentum along X - axis is still zero.

Therefore we can say,

m1.v1(x) + m2v2(x) = 0

94.8 * 5.43 + 57.7 * v2(x) = 0

57.7 * v2(x) = -512.592

v2(x) = -512.592 / 57.7

v2(x) = -8.88 m/s

If we also consider the momentum along Y - axis:

Before they got separated, the momentum = 0

After they got separated, the momentum along Y - axis is still 0

Therefore we can say that

m1v1(y) + m2v2(y) = 0

94.8 * 4.25 + 57.7 * v2(y) = 0

57.7 * v2(y) = -402.9

v2(y) = -402.9 / 57.7

v2(y) = -6.98 m/s

Therefore, the magnitude of the x and y component of the velocity are

-8.88 m/s and -6.98 m/s respectively.

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this