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## Main Question or Discussion Point

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?

Here is what I did to try and get the answer.. First I found Voy by using: 2.05(sin35) which is 1.17m, then I found Vy by using.. -g(t) which is -9.80 ( 1.60)=-15.68.. using these values I plugged them into the H equation which is: H=Vy^2-Voy^2/2Ay... and I got 12.5 for an answer which is wrong.. can anyone tell me where I went wrong.. it would be much appreciated.. thanks

Here is what I did to try and get the answer.. First I found Voy by using: 2.05(sin35) which is 1.17m, then I found Vy by using.. -g(t) which is -9.80 ( 1.60)=-15.68.. using these values I plugged them into the H equation which is: H=Vy^2-Voy^2/2Ay... and I got 12.5 for an answer which is wrong.. can anyone tell me where I went wrong.. it would be much appreciated.. thanks