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

[New-Blu-Blood]

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.

 

[rl.bhat]

Show your calculations. I am getting different answer.

 

[New-Blu-Blood]

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

 

[rl.bhat]

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.