Kinematics / forces - Young girl swinging on a rope

[Huskies213]

On a hot summer day a young girl swings on a rope above the local swimming hole. When she let's go of the rope her initial velocity is 2.65 m/s at an angle of 35.0° above the horizontal. If she is in flight for 1.30 s, how high above the water was she when she let go of the rope?

can anyone help?? thanks.

 

[KD]

To solve this problem, it is best to set up a chart and draw a picture.
We already know the time is 1.30s and the accleration is 9.81 m/s2 - that's always a given. To get the vertical height, you need a vertical initial velocity. You need to break the velocity they give you into components. Make a right triangle and put that 2.65 at the hypotenuse. Solve for the vertical inital velocity by 2.65sin35.
After that you can plug it into the equation d=vi t +.5 a t^2
This will give you the height.

 

[lightgrav]

First separate the velocity vector into vertical and horizontal components.
Then use the (known) vertical acceleration and the (known) time
to compute the change in vertical location.