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On a hot summer day, a young girl swings on a rope above the local swimming hole (Figure 1) . When she lets go of the rope her initial velocity is 2.10m/sm/s at an angle of 35.0 ∘^\circ above the horizontal.
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|dw:1410132889685:dw|
i dont think you completed the question
t= 0.618s
i mean, what is the question asking for?
how high above the water was she when she go off the rope?
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hmm can you post the figure?
hmm ok i guess u know this equation, \[y=(v _{0}\sin \theta)t-1/2 g t^2\] you need to find out "y" @yuhan
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