Asked by Mark
An aerialist on a high platform holds onto a trapeze attached to a support by an 8.0-m cord. (See the drawing.) Just before he jumps off the platform, the cord makes an angle of 41° with the vertical. He jumps, swings down, then back up, releasing the trapeze at the instant it is 0.75 m below its initial height. Calculate the angle θ that the trapeze cord makes with the vertical at this instant.
Answers
Answered by
Elena
Imagine two triangles:
1) ΔABC : point A (where the cord is attached ), point B (where the aerialist holds the trapeze at initial position), point C (the intersection of the vertical line from A and horizontal line from B),
2) ΔADM: point M (where the aerialist holds the trapeze at its left position), point D (the intersection of the vertical line from A and horizontal line from M)
AC = AB•cosθₒ =L•cosθₒ,
AD =AM•cosθ = L•cosθ,
CD = d.
AD =AC +CD,
L•cosθ = L•cosθₒ +d.
cosθ =( L•cosθₒ +d)/L =
=(8•cos41º+0.75)/8 = 0.848.
arcos 0.848 = 32º.
1) ΔABC : point A (where the cord is attached ), point B (where the aerialist holds the trapeze at initial position), point C (the intersection of the vertical line from A and horizontal line from B),
2) ΔADM: point M (where the aerialist holds the trapeze at its left position), point D (the intersection of the vertical line from A and horizontal line from M)
AC = AB•cosθₒ =L•cosθₒ,
AD =AM•cosθ = L•cosθ,
CD = d.
AD =AC +CD,
L•cosθ = L•cosθₒ +d.
cosθ =( L•cosθₒ +d)/L =
=(8•cos41º+0.75)/8 = 0.848.
arcos 0.848 = 32º.
Answered by
Elena
arccos 0.848 = 32º.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.