Solving Kinematics: Swinging Girl Above Swimming Hole

[Mesmer17]

Q.)On a hot summer day a young girl swings on a rope above the local swimming hole (Figure 4-20). When she let's 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

 

[LightHero]

!!Your equation is not correct. The equation for the height is = Voy*t - 0.5*g*t^2Substituting the values you have, the equation becomes = 1.17*1.60 - 0.5*(-9.8)*(1.60)^2This gives a height of 7.4 meters.