The problem describes a scenario involving a container with water jets emanating from it at different heights. The goal is to analyze the water jet trajectories and understand the physics behind them. Specifically, parts (f), (g), and (h) ask to calculate the horizontal distance AC where the water jet hits the ground, determine the point B where the jet travels the furthest, and describe the relationship between water jet range and depth.

1. Problem Description

The problem describes a scenario involving a container with water jets emanating from it at different heights. The goal is to analyze the water jet trajectories and understand the physics behind them. Specifically, parts (f), (g), and (h) ask to calculate the horizontal distance AC where the water jet hits the ground, determine the point B where the jet travels the furthest, and describe the relationship between water jet range and depth.

2. Solution Steps

Let's address the calculations related to parts (f), (g), and (h). We are given $h = 25 cm$, $H = 100 cm$, $\rho = 1000 kg/m^3$, and $g = 10 m/s^2$.

(f) Find the horizontal distance $AC$ for the bottom hole.

The velocity of the water jet emerging from the hole at height $h$ above the bottom is given by Torricelli's theorem:

$v = \sqrt{2g(H-h)}$

The time it takes for the water jet to hit the ground is:

$t = \sqrt{\frac{2h}{g}}$

The horizontal distance $AC$ is the product of the velocity and time:

$AC = v \cdot t = \sqrt{2g(H-h)} \cdot \sqrt{\frac{2h}{g}} = \sqrt{4h(H-h)} = 2\sqrt{h(H-h)}$

Converting the given values to meters: $h = 0.25 m$ and $H = 1.00 m$.

$AC = 2\sqrt{0.25(1-0.25)} = 2\sqrt{0.25(0.75)} = 2\sqrt{0.1875} = 2(0.433) \approx 0.866 m$

(g) Find the depth where the horizontal distance is maximized.

Let the depth of the hole from the surface of the water be $x$. Therefore, the height of the hole above the ground is $H-x$. Then, the horizontal distance will be:

$d = 2\sqrt{(H-x)x} = 2\sqrt{Hx - x^2}$

To maximize the horizontal distance, we need to maximize the expression inside the square root. Let's find the value of $x$ that maximizes the quadratic $Hx - x^2$.

Taking the derivative with respect to $x$ and setting it to zero:

$\frac{d}{dx}(Hx - x^2) = H - 2x = 0$

$x = \frac{H}{2}$

Therefore, the maximum horizontal distance occurs when the hole is at a depth of half the total height.

The depth at B is $x = \frac{H}{2} = \frac{100cm}{2} = 50cm$

So, B is $50cm$ from the surface.

(h) The range of the water jet depends on the depth of the hole from the surface. When the hole is near the surface or near the bottom, the range is small. The range is maximized when the hole is at half the depth.

3. Final Answer

(f) The horizontal distance $AC$ is approximately $0.866 m$.

(g) The water jet travels furthest when the hole is at a depth of $50 cm$.

(h) The range increases from zero at the surface to a maximum at half the depth and then decreases back to zero at the bottom.