An aerialist on a high platform holds onto a trapeze attached to a support by an 9.2-m cord. (See the drawing.) Just before he jumps off the platform, the cord makes an angle of 44° with the vertical. He jumps, swings down, then back up, releasing the trapeze at the instant it is 0.54 m below its initial height. Calculate the angle θ that the trapeze cord makes with the vertical at this instant.
Solution,
Y1 = 9.2 cos(44) = 6.62
Y2 = 6.62 + 0.54 = 7.16
9.2 cos(theta) = 7.16
Cos (theta) = 7.16/9.2
Theta = 38.92 degree
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