A child whose weight is 357 N swings out over a pool of water using a rope attac

Question

A child whose weight is 357 N swings out over a pool of water using a rope attached to the branch of a tree at the edge of the pool. The branch is 12.5 m above ground level and the surface of the water is 1.77 m below ground level. The child holds on to the rope at a point 10.4 m from the branch and moves back until the angle between the rope and the vertical is 22.8 degrees;. When the rope is in the vertical position, the child lets go and drops into the pool. Calculate the speed of the child just as he impacts the surface of the water.

I calculated the maximum velocity to be 3.9909 m/s in the x direction from sqrt(2(9.8m/s^2)(10.4m)(1-cos22.8)) and the velocity in the y direction to be 5.89 m/s (v^2=2(9.8 m/s^2)(1.77 m), but I'm can't figure out how to combine the two (the Pythagorean theorem did not give me the right answer) or if I'm just on the completely wrong track.

Explanation / Answer

         The weight of the child, W = 357 N          The distance between bracnch of the tree and          surface of the water, h = 12.5 m + 1.77 m                                              = 14.27 m          The height traveled by the child with an          angle 22.8o (between the rope and vertical) is                  h' = (10.4 m)cos22.8o _________________________________________________________________       Using the law of conservation of energy,                       (K.E)i+(P.E)i = (K.E)f+(P.E)f                        0 + mg(h-h') = (1/2)mv2                                 g(h-h') = (1/2)v2       Then, the speed of the child just as he impacts the surface of the water is                           v = 2g(h-h')                               = (2)(9.8 m/s2){(14.27 m-(10.4 m)cos22.8o)}                               = 9.58 m/s                               = 9.6 m/s       Then, the speed of the child just as he impacts the surface of the water is                           v = 2g(h-h')                               = (2)(9.8 m/s2){(14.27 m-(10.4 m)cos22.8o)}                               = 9.58 m/s                               = 9.6 m/s

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