Two acrobats, each of 64.0 kg, launch themselves together from a swing, holding

Question

Two acrobats, each of 64.0 kg, launch themselves together from a swing, holding hands. Their velocity at launch was 5 60 m's and the angle of their initial velocity relative to the horizontal was 720 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?

Explanation / Answer

mass of each acrobat, m = 64 kg
veolcity at launch, u = 5.6 m/s
phi = 72 deg

at the top of the trajectory
horizontal distance from the original swing = ucos(phi)*t
where
usin(phi) = gt
hence
d1 = u^2cos(phi)sin(phi)/g .. (1)
height h = usin(phi)*t - 0.5gt^2 = u^2sin^2(phi)/g - u^2*sin^2(phi)/2g = u^2*sin^2(phi)/2g
at this point, momentum of system = 2mucos(phi)
so as one acrobat comes to rest speed of other = v
from conservation of momentum
2mucos(phi) = mv
v = 2ucos(phi)

time taken to fall from this height = t = usin(phi)/g
hence total horizontal range = d1 + 2ucos(phi)usin(phi)/g
d = u^2cos(phi)sin(phi)/g + 2u^2sin(phi)cos(phi)/g = 3u^2sin(phi)cos(phi)/g = 3*5.6^2*sin(72)cos(72)/g
d = 2.8184 m

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