Optimal Launch Speed for Swinging Across a Ravine

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SUMMARY

The discussion focuses on calculating the minimum horizontal speed required for a hiker to swing across a ravine using a rope. The parameters given are L = 4.0 m and x = 1.8 m. The key equations utilized include the conservation of energy, represented as 1/2mv² = mgh, and the relationship between the lengths of the rope and the height using trigonometric functions. The confusion arises regarding the correct method to determine the angle and height, with suggestions to use the Pythagorean theorem for height calculation instead of trigonometric functions.

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This discussion is beneficial for physics students, educators, and anyone interested in solving mechanics problems involving energy conservation and trigonometry.

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Homework Statement


A hiker plans to swing on a rope across a ravine in the mountains, as illustrated in the figure, where L = 4.0 m and x = 1.8 m, and to drop when she is just above the far edge. At what minimum horizontal speed should she be moving when she starts to swing(in m/s)?

TNDhEC6.gif


Homework Equations



Ei= Ef
So...
Ke = Pe
So...
1/2mv2=mgh

The Attempt at a Solution



I have the equation set up correctly but I just don't know how to find h so I can find V

1/2mv2=mgh

The Rope makes a isosceles triangle shape so I thought of L as adjacent, and X as the Opposite so I could solve for theta by doing Θ = arctan(1.8/4.0) = 24.22º. However, upon looking at the answer online- it says to take the arcsin(1.8/4.0) = 26.7º. Why is this?
 
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##L## is not the side adjacent in the right triangle.
 

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For that matter, x is not the base of an isosceles triangle, as TSny's diagram illustrates.
That is not the formula for finding the apex angle of an isosceles triangle anyway.
As TSny says: ##x\neq L\tan\theta## either.

I'm kinda puzzled that they want you to find the angle at all.
Since you know L and x, why not find h by pythagoras?
 

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