Two acrobats, each of 60.0 kg, launch themselves together from a swing holdings

Question

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?

Explanation / Answer

at the top of their arc, they have no vertical motion, so have only horizontal motion of 8cos46 m/s=5.55m/s,,,,,,,,, the total momentum before pushing (horizontally) was 2*60*(8 cos 46) =666,,,,,,,,,,,,,, calculate the time to reach max height (8 sin 46/g)=0.586 s,,,,,,,,,,,,,,,,,,the two swings should be 3/2 the range, or 3/2 x v0^2 sin(2*theta)/g = 1.5*64*sin(92)/9.81=9.78

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