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 2.25m/s at an angle of 35.0 above the horizontal. If she is in flight for 0.616sec, how high above the water was she when she let go of the rope?

Answer 1

Her vertical velocity component at release was 2.25*sin 35 = 1.29 m/s

Solve the following equation for altitude vs. time to get the initial height H:

y = H + 1.29 t - (g/2) t^2
When t = 0.616 s, y = 0
Use that to solve for H

Answer 2

1m

Answer 3

What is g stand for