Two acrobats, each of 62.0 kg, launch themselves together from a swing, holding

Question

Two acrobats, each of 62.0 kg, launch themselves together from a swing, holding hands. Their velocity at launch was 6.40 m/s and the angle of their initial velocity relative to the horizontal was 44.0 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

Their vertical velocity at launch is:

v_oy = 6.4*Sin(44) = 4.44m/s

v_ox = 6.4*Cos(44) = 4.60m/s

The time to the top of the trajectory is:

v_y = 0 = v_oy -g*t

v_oy = g*t

t = v_oy/g = 0.4526 seconds

In this time, they have traveled:

x = v_ox*t = 4.60* 0.4526 = 2.08m

At the top of the trajectory, all of the velocity and therefore all of the energy is in the "x" direction. Since energy is conserved, and one acrobat comes to rest, the energy is transferred to the fellow performer.

KE_2 = 2*KE_1

(1/2)*m*v^2 = 2*(1/2)m*(v_ox)^2

the (1/2) cancels, the "m" cancels,

v^2 = 2*(v_ox)^2

v = sqrt(2)*(v_ox) = sqrt(2)*4.6 = 6.50m/s

The "y" velocity is unaffected by the push, and so the time required to descend back to the take-off level is the same, the distance covered in "x" becomes:

x_2 = 6.50*0.4526 = 2.94m

The total x distance covered is:

x = 2.08 + 2.94 = 5.02m

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