It's a basic freefall equation:
First the total time they are in the air will remain the same because the only acceleration will be gravity. So you can calculate time with 0=(6.8sin(36))t-4.9t^2, as you would for any free fall.
The horizontal velocity will tell us how far apart the two swings. As such, the horizontal distance can be split into two parts, before they push off (v1) and after (v2).
v1 will be 6.8cos(36).
Since v2 is the velocity needed to negate one of the acrobats speed, (they're traveling at the same speed) v2=2(v1)
They push off at the vertex, so the time of each segment would be t/2.
So, the first distance traveled would be (v1(t/2)). The second segment would therefore be ((v2)(t/2)) or (2(v1)(t/2)). add them together and you get 3(v1)(t/2)=x
Sorry, no calculator at the moment. hope this helps
Two acrobats, each of 50.0 kg, launch themselves together from a swing, holding hands. Their velocity at launch was 6.80 m/s and the angle of their initial velocity relative to the horizontal was 36.0 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 have no idea how to even start this question......
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