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?

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)

To determine the horizontal distance traveled by water passing through a hole in a punctured water bottle, we can utilize Bernoulli's equation and kinematics.

Let's start by considering the situation. We have a water bottle with water inside, where the height of the water is represented by "y." The bottle has a hole located at a height "h" from the floor. We want to find the horizontal distance traveled by the water after it passes through the hole.

First, let's analyze the forces acting on the water inside the bottle. We can assume that the only significant forces are the pressure force and the weight of the water. Bernoulli's equation relates the pressure, velocity, and height of a fluid in a horizontal streamline.

Bernoulli's equation can be written as:
P + 1/2 * ρ * v^2 + ρ * g * y = constant

Where:
P is the pressure of the fluid
ρ is the density of the fluid (in this case, water)
v is the velocity of the fluid
g is the acceleration due to gravity
y is the height of the fluid

Now, let's consider two points in the system: the surface of the water inside the bottle and the point where the water exits the hole. At the surface of the water, the velocity is approximately zero (assuming a still body of water). At the exit point, the velocity can be assumed to be proportional to the square root of the depth of the water in the bottle. Therefore, we can write:

v = k * sqrt(y)

By applying Bernoulli's equation at these two points, we can derive an equation for the velocity at the exit point:

P + 1/2 * ρ * 0^2 + ρ * g * y(surface) = P + 1/2 * ρ * (k * sqrt(y))^2 + ρ * g * y(exit)
0 + 0 + ρ * g * y(surface) = P + 1/2 * ρ * k^2 * y + ρ * g * y(exit)
ρ * g * y(surface) = P + 1/2 * ρ * k^2 * y + ρ * g * y(exit)

Next, let's consider the horizontal distance traveled by the water. We can use kinematics to determine the displacement along the horizontal direction. We assume that the acceleration in this direction is zero, and the initial velocity is zero. Therefore, we have:

x = V(initial) * t + 1/2 * a * t^2
Since V(initial) = 0 and a = 0, x = 0

This means that the water will travel horizontally from the hole without any deviation due to the absence of horizontal forces acting on it. Therefore, the horizontal distance traveled, delta x, is equal to zero.

In conclusion, given the setup and assumptions outlined above, the horizontal distance traveled by the water passing through the hole will be zero.