Optimal Launch Speed for Swinging Across a Ravine

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To determine the minimum horizontal speed for a hiker swinging across a ravine, the key equations involve conservation of energy, specifically 1/2mv² = mgh. The challenge arises in calculating the height (h) from the given lengths L and x. There is confusion regarding the correct angle to use for calculations, with some suggesting to use arcsin instead of arctan for the triangle formed. It is also noted that using the Pythagorean theorem may be a more straightforward approach to find h. Ultimately, clarity on the triangle's geometry is essential for solving the problem correctly.
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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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