Question

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 35.0° 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?

Answer 1

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)

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