Asked by paul
the height of the upper falls at Yellowstone falls is 33.2m. when the water reaches the botton of the falls, its speed is 25.8 m/s. neglecting air resistance, what is the speed of the water at the to of te falls.
Answers
Answered by
drwls
The water acquires a vertical velocity component during the fall equal to
Vy = sqrt(2gH) = 25.5 m/s. Its velocity at the bottom is 25.8 m/s, so the horizontal component at the bottom must be sqrt[(25.8)^2 - (25.5)^2] = 3.9 m/s
That must be the horizontal component at the top also, since it does not change during the fall. It is the velocity magnitude there.
Vy = sqrt(2gH) = 25.5 m/s. Its velocity at the bottom is 25.8 m/s, so the horizontal component at the bottom must be sqrt[(25.8)^2 - (25.5)^2] = 3.9 m/s
That must be the horizontal component at the top also, since it does not change during the fall. It is the velocity magnitude there.
Answered by
Rose
3.9
Answered by
Yeah
Yeah
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