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?

Question

Physics

Bryson Yoder

1 July, 20:39

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?

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Answers (1)

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Jaylen Stokes · 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.