Two acrobats, each of 58.0 kg, launch themselves together from a swing, holding

Question

Two acrobats, each of 58.0 kg, launch themselves together from a swing, holding hands. Their velocity at launch was 5.80 m/s and the angle of their initial velocity relative to the horizontal was 54.0 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?

I plugged all these numbers (9.40, 2.77, 3.25 and 5.09) but none of them was correct.

Explanation / Answer

at the top they have only horizontal motion of 5.8cos54 m/s = 3.40 m/s

since there are no horizontal forces acting on the system, the total horizontal momentum is conserved

so initial momentum = 2 *m* (5.8 cos 54) = 395.46 kg m/s

we could solve this as a center of mass problem, knowing the motion of the center of mass is unchanged

if there were no pushing involved, they would each travel a distance D given by the range equation

and this will be the distance traveled by the center of mass; one person will only travel a distance D/2, so the other person must travel 3D/2 so that the center of mass travels a distance D

therefore, the two swings should be 3/2 the range,

now range is given by v2sin(2*theta) /g

or 3/2 x    v2 *sin(2*theta)/g =4.891m/s

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