Question

Two skydivers are holding on to each other while falling straight down at a common terminal speed of 60.70 m/s. 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):

Answer 1

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) :

v_1x = 4.43 m/s , v_1y = 5.25 m/s , v_1z = 63.1 m/s

(a) The x & y components of the velocity of the second skydiver, whose mass is 52.2 kg, immediately after separation which is given as :

let m₁ be the mass of the first skydiver = 89.3 kg

m₂ be the mass of the second skydiver = 52.2 kg

When they fall together, it means that -

M_x = 0, M_y = 0

When they push away from each other, it means that -

M_x = m₁ v_x1 - m₂ v_x2 = 0                                                                

v_x2 = m₁ v_x1 / m₂                                                                    { eq.1 }

inserting the values in eq.1,

v_x2 = (89.3 kg) (4.43 m/s) / (52.2 kg)

v_x2 = 7.57 m/s

AND

M_y = m₁ v_y1 - m₂ v_y2 = 0

v_y2 = m₁ v_y1 / m2                                                                             { eq.2 }

inserting the values in eq.2,

v_y2 = (89.3 kg) (5.25 m/s) / (52.2 kg)

v_y2 = 8.98 m/s