How High Was the Girl Above the Water When She Released the Rope?

AI Thread Summary
The problem involves calculating the height above water when a girl releases a rope while swinging. The initial velocity is 2.10 m/s at a 35-degree angle, and she is in flight for 0.615 seconds. The equation used is y = (v0sin(theta))t - 0.5gt^2, but there is confusion regarding the sign of g. The initial calculations yielded incorrect heights, with the book stating 1.07 meters as the correct answer. Proper application of the equation, considering the direction of forces, is crucial for accurate results.
New-Blu-Blood
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Homework Statement


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 2.10 m/s at an angle of 35.0 degrees above the horizontal. If she is in flight for 0.615 s, how high above the water was she when she let go of the rope?


Homework Equations



The equation I used to try to solve for the problem was y=(v0sin(theta))t-.5gt2

The Attempt at a Solution



I know the angle is 35 degrees, initial velocity is 2.10 m/s and that initial X and Y positions are both zero. I plug my numbers into the equation given above and get 3.15 for y. The answer in the back of the book says 1.07 meters, but I can't figure out how they arrived at this answer.
 
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Show your calculations. I am getting different answer.
 
Ok, I think I messed the calculation up a little. Here should be the correct calculation.

y= (2.10 x sin 35).615 - .5(-9.80).6152 = 2.59 or 3
 
New-Blu-Blood said:
Ok, I think I messed the calculation up a little. Here should be the correct calculation.

y= (2.10 x sin 35).615 - .5(-9.80).6152 = 2.59 or 3
When you use the formula
y = vo*t - 0.5*g*t^2, you assume that y and vo are in the upward direction and g is in the downward direction. So you should not use -g in the equation.
 
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