Two acrobats, each of 60.0 kg, launch themselves together from a swing holdings hands. Their velocity at launch was 8.00 m/s and the angle of their initial velocity relative to the horizontal was 46.0 degrees upwards. At the top of their trajectory, their act calls for them to push against each in such a way that one of them becomes stationary in mid-air and then falls to the safety net while the other speeds away, to reach another swing at the same height at that of the swing they left. What should be the distance (in m) between the two swings?

To solve this problem, we need to apply the conservation of momentum and the principle of conservation of mechanical energy.

Step 1: Calculate the initial total momentum of the system.
The initial total momentum of the system will be the sum of the individual momenta of the acrobats. Since they are holding hands and launch themselves together, their initial momenta will have the same magnitude but opposite in direction.

Given:
Mass of each acrobat, m = 60.0 kg
Velocity of each acrobat, v = 8.00 m/s

The initial momentum of one acrobat = mass x velocity = 60.0 kg x 8.00 m/s = 480 kg.m/s
Since the other acrobat's momentum is in the opposite direction, its momentum will be -480 kg.m/s.

Therefore, the initial total momentum of the system = 480 + (-480) = 0 kg.m/s

Step 2: Calculate the change in momentum for the acrobat who becomes stationary.

At the top of their trajectory, the stationary acrobat will have a momentum of zero, so the change in momentum is equal to the initial momentum of that acrobat.

Change in momentum = 480 kg.m/s

Step 3: Apply the principle of conservation of mechanical energy.

At the top of the trajectory, the acrobats' potential energy is maximum, and their kinetic energy is zero. Therefore, we can equate the initial potential energy to the final potential energy.

Potential Energy at the top = mgh
Initial potential energy = 2 x m x g x h (Since there are two acrobats)

Final potential energy = mgh (Since one acrobat falls to the safety net)

The mass, m, and the acceleration due to gravity, g, are the same in both cases. Therefore, we only need to consider the difference in height, h, between the swings.

Step 4: Calculate the change in height.

Since the acrobat who falls to the safety net becomes stationary in mid-air, we know that his velocity at that point is zero. Using the equation for final velocity in free-fall motion:

Final velocity, vf = 0 m/s
Initial velocity, vi = 8.00 m/s
Acceleration due to gravity, g = 9.8 m/s^2

vf^2 = vi^2 + 2ad

Solving for d (the displacement):

0^2 = 8.00^2 + 2 * 9.8 * d

64 = 19.6d

d = 64 / 19.6 = 3.27 m

Step 5: Solve for the distance between the swings.

Since the height difference between the swings is equal to the change in height:

Change in height, h = 3.27 m

Using the principle of conservation of mechanical energy:

2 x m x g x h = m x g x h

Simplifying,

2 x 60.0 kg x 9.8 m/s^2 x h = 60.0 kg x 9.8 m/s^2 x 3.27 m

h = 3.27 / 2 = 1.635 m

Therefore, the distance between the two swings should be approximately 1.635 meters.