How High Was the Girl When She Released the Rope?

[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?

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

 

[jenny10]

Kinematics

Use the equation Vy= Voy + at to find Vy. You only multiplied a by t, so your Vy term was incorrect. Good luck

 

[HallsofIvy]

Since you are told that "she is in flight for 1.60 s", don't you think it would be a good idea to use that information?

The simplest way to do this problem is to note that the height at time t is H= -g/2 t²+ v0 sin(theta) t+ y0 where H is the height above the water and y0 is the initial height above the water (which is what you are asked for).

When she hits the water, H= 0 so plug in the information you are given and solve -g/2 t²+ vo sin(theta)+ y0= 0 for y0.