Asked by AfterLife
How do I determine the horizontal distance traveled by water passing through a hole of a punctured water bottle. You must use bernoulli's and Kinematics, and find delta x in term of h and y. ( y = the height of water inside the bottle , h = the height of the hole to the floor )
My attempt bernoulli's equation says P+ 1/2 rv^2+ rgy=constant , kinematics says Vf^2 = Vi^2 + 2ax
therefore, combining them together ,, 1/2 r( Vi + sqrt 2ax )^2 + rgy.
But I think it is wrong , as I am suppose to find x in terms of h :S Help please , how can I find it?
My attempt bernoulli's equation says P+ 1/2 rv^2+ rgy=constant , kinematics says Vf^2 = Vi^2 + 2ax
therefore, combining them together ,, 1/2 r( Vi + sqrt 2ax )^2 + rgy.
But I think it is wrong , as I am suppose to find x in terms of h :S Help please , how can I find it?
Answers
Answered by
drwls
The speed of the water leabving the hole is
V = sqrt[2 g (y-h)]
That comes from using the Bernoulli equation. y-h is the height of the water above the hole. Multiply V by the time that it takes the water to hit the ground, to get the horizontal distance travelled.
(1/2) g t^2 = y
t = sqrt(2y/g)
V = sqrt[2 g (y-h)]
That comes from using the Bernoulli equation. y-h is the height of the water above the hole. Multiply V by the time that it takes the water to hit the ground, to get the horizontal distance travelled.
(1/2) g t^2 = y
t = sqrt(2y/g)
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