Kinematics / forces - Young girl swinging on a rope

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SUMMARY

The discussion focuses on calculating the height a young girl reaches when she lets go of a rope while swinging. Given her initial velocity of 2.65 m/s at an angle of 35.0° and a flight time of 1.30 seconds, the vertical component of her initial velocity is determined using the formula 2.65sin(35°). The vertical height is then calculated using the equation d = vi * t + 0.5 * a * t², where acceleration due to gravity is 9.81 m/s². This method provides a clear approach to solving projectile motion problems involving initial velocity and time of flight.

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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.
 
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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 acceleration 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.
 
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.
 

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