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 veloci
ty 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?
1 answer:
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.
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.
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