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 89.30 kg) has the following velocity components (with ''straight down'' corresponding to the positive z-axis): v1,x=4.430 m/s v1,y=4.250 m/s v1,z = 52.30 m/s What are the x- and y-components of the velocity of the second skydiver, whose mass is 63.20 kg, immediately after separation?

Answer 1

Answer:

$v_{2,x} = - 6.259\ m/s$

$v_{2,y} = - 6.005\ m/s$

Solution:

As per the question:

Common terminal speed, $v_{cT} = 52.30\ m/s$

Mass of the one of the skydiver, $m_{1}$ = 89.30 kg

Velocity of the skydiver :

$v_{1,x} = 4.430\ m/s$

$v_{1,y} = 4.250\ m/s$

$v_{1,z} = 52.30\ m/s$

Mass of the other skydiver, $m_{2}$ = 63.20 kg

Now,

To calculate the components of velocity along X and Y axes:

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

After the separation, the momentum along X-axis is zero.

Therefore,

$m_{1}v_{1,x} + m_{2}v_{2,x} = 0$

$89.30\times 4.430 + 63.20\times v_{2,x} = 0$

$v_{2,x} = - 6.259\ m/s$

Now, consider the momentum along Y-axis:

Before separation, momentum = 0

After separation, momentum along Y-axis = 0

Therefore,

$m_{1}v_{1,y} + m_{2}v_{2,y} = 0$

$89.30\times 4.250 + 63.20\times v_{2,y} = 0$

$v_{2,y} = - 6.005\ m/s$

Thus the magnitude of the X and Y component of velocity are 6.259 m/s and 6.05 m/s respectively.