HELP ! 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 find the distance between the two swings, we need to analyze the motion of the acrobats. We can break down their motion into two parts: the projectile motion after launch and the relative motion when they push against each other.

First, let's calculate the initial vertical and horizontal components of their velocity at launch.

Vertical velocity (vy):
vy = velocity * sin(angle)
vy = 8.00 m/s * sin(46.0 degrees)
vy = 8.00 m/s * 0.7193
vy = 5.7544 m/s

Horizontal velocity (vx):
vx = velocity * cos(angle)
vx = 8.00 m/s * cos(46.0 degrees)
vx = 8.00 m/s * 0.6947
vx = 5.5576 m/s

Next, we need to determine the time it takes for one of the acrobats to become stationary at the top of their trajectory. At the highest point, the vertical velocity becomes zero (vy = 0). We can use vertical motion equations to find the time it takes.

vy = vy_initial + (-g * t)
0 = 5.7544 m/s - 9.8 m/s^2 * t

Solving for t:
9.8 m/s^2 * t = 5.7544 m/s
t = 5.7544 m/s / 9.8 m/s^2
t = 0.5872 s

Now, we can determine the maximum height reached by the acrobats, which is the distance between the swings. We can use the vertical motion equation:

y = y_initial + vy_initial * t + (0.5 * (-g) * t^2)
y = 0 + 5.7544 m/s * 0.5872 s + (0.5 * (-9.8 m/s^2) * (0.5872 s)^2)

y = 1.6909 m

Therefore, the distance between the two swings should be approximately 1.6909 meters.