Q)On a hot summer day a young girl swings on a rope above the local swimming hole (Figure 4-20). When she lets go of the rope her initial velocity is 2.05 m/s at an angle of 35.0° above the horizontal. If she is in flight for 1.60 s, how high above the water was she when she let go of the rope?(adsbygoogle = window.adsbygoogle || []).push({});

TO GET THE ANSWER.. first I found Vo by using 2.05(sin35) which is 1.17.. then I got the Vy by using -G(t) which is -9.80(1.60) which I got 15.68 for.. using these values I plugged them into the equation: H=Vy^2-Voy^2/2Ay and I got 12.5 for an answer that is wrong.. can anyone tell me where I went wrong? THANKS

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# Homework Help: Kinematics rope question

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