Thread: General Launch Angle Question

1. The problem statement, all variables and given/known data On a hot summer day, a young girl swings on a rope above the local swimming hole . When she lets 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? 2. Relevant equations The equation I used to try to solve for the problem was y=(v 0...

Originally Posted by 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) .615 2 = 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.