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 2.25 m/s at an angle of 35.0â° above the horizontal. if she is in flight for 1.10 s, how high above the water was she when she let go of the rope?

Refer to the diagram shown below.
 h =  height of the girl above water when she lets go of the rope.

The launch velocity is 22.5 m/s at 35° to the horizontal. Therefore the vertical component of the velocity is
v = 22.5 sin(35°) = 12.9055 m/s.

The time of flight is t = 1.10 s before the girl hits the surface of the water at a height of  -h.
Therefore
-h = (12.9055 m/s)*(1.10 s) - (1/2)*(9.8 m/s²)*(1.10 s)²
-h = 8.267 m
    = 8.3 m (nearest tenth)

 
When the girl let go of the rope, she was about 8.3 m  above the surface of the water.