Water flows out in all directions with the same speed from a sprinkler consisting of a perforated sp

Q: Why do we assume the maximum height is reached when the angle is 90 degrees?

The formula for maximum height is $H = \frac{v^2 \sin^2\theta}{2g}$. Since $\sin\theta$ is maximum at $90^\circ$ (where it equals 1), the water shot straight up will reach the highest point.

Q: Why is the watered area a circle?

The sprinkler has a spherical shell and shoots water in all directions symmetrically. The maximum horizontal range is the same in every direction, forming a perfect circle on the ground.

Q: Can I just use the standard range formula $R = \frac{v^2 \sin 2\theta}{g}$ for the second case?

No, that formula is only valid when the projectile lands at the same vertical level it was launched from. Since the sprinkler is elevated, you must use the trajectory equation or the specific formula for range from a height.

Q: What is the "envelope of trajectories"?

It's a boundary curve that encloses all possible paths the water can take for a given speed. Any point inside this envelope can be hit by the water, and no point outside it can be reached. It's a powerful tool for finding maximum range!

Q: Does the angle for maximum range change when the sprinkler is elevated?

Yes! On the ground, the maximum range occurs at $45^\circ$. When elevated, the optimal angle is shallower (less than $45^\circ$) because the projectile spends extra time falling, allowing a more horizontal velocity to cover more ground.

Question

Water flows out in all directions with the same speed from a sprinkler consisting of a perforated spherical shell fixed at the end of a hose. When the sprinkler is fixed at the ground, maximum height attained by a water stream is $h$. If the sprinkler is shifted to height $h$ above the ground, by what factor will the watered area on the ground change? Neglect diameter of the spherical shell as compared to the height $h$.

Sigma Percentile · Accepted Answer

The problem asks us to find how the watered area changes when a spherical sprinkler is elevated from the ground to a height $h$. This is a classic application of projectile motion, but with a twist: we are dealing with projectiles fired in *all* directions simultaneously.

### Visualizing the Sprinkler's Reach

Imagine the sprinkler sitting on the ground. It's a perforated sphere, meaning water shoots out at every possible angle with a constant speed $v$. The water droplet that is fired straight up ($	heta = 90^\circ$) will reach the maximum possible vertical height. The problem states this maximum height is $h$.

From the kinematics of a projectile, the maximum height is given by:
$$H = \frac{v^2 \sin^2	heta}{2g}$$

For the vertically fired droplet, $\sin 90^\circ = 1$, so:
$$h = \frac{v^2}{2g}$$

This simple relation allows us to find the ejection speed of the water:
$$v = \sqrt{2gh}$$

This speed $v$ is the same for every droplet leaving the sprinkler, regardless of its launch angle.

### The Ground Level Setup

To find the area watered by the sprinkler when it's on the ground, we need to determine how far the water can travel horizontally. The maximum horizontal range $R_1$ on level ground occurs at a launch angle of $45^\circ$ and is given by:
$$R_1 = \frac{v^2}{g}$$

Substituting our expression for $v^2$:
$$R_1 = \frac{2gh}{g} = 2h$$

Because the sprinkler shoots water symmetrically in all directions, the watered region is a perfect circle with radius $R_1$. The initial watered area $A_1$ is:
$$A_1 = \pi R_1^2 = \pi (2h)^2 = 4\pi h^2$$

### Elevating the Sprinkler

Now, we elevate the sprinkler to a height $h$ above the ground. The water still exits with the same speed $v = \sqrt{2gh}$, but because it starts higher, it will spend more time in the air and thus travel further horizontally.

We need to find the new maximum horizontal range, $R_2$. While we could use the standard formula for the range of a projectile fired from a height, a more elegant approach is to use the **envelope of trajectories**.

### The Envelope of Trajectories

The envelope of trajectories represents the absolute boundary of all possible parabolic paths the water can take. For a projectile fired from the origin $(0,0)$ with speed $v$, the bounding parabola is:
$$y = \frac{v^2}{2g} - \frac{g x^2}{2v^2}$$

Since our sprinkler is now at a height $h$, we shift the origin up by $h$. The new envelope equation becomes:
$$y = h + \frac{v^2}{2g} - \frac{g x^2}{2v^2}$$

Let's substitute $v^2 = 2gh$ into this equation to simplify it:
$$y = h + \frac{2gh}{2g} - \frac{g x^2}{2(2gh)}$$
$$y = h + h - \frac{x^2}{4h}$$
$$y = 2h - \frac{x^2}{4h}$$

This beautiful equation describes the outer limit of the water spray. To find the maximum range $R_2$ on the ground, we set $y = 0$ (the ground level) and solve for $x$:
$$0 = 2h - \frac{R_2^2}{4h}$$
$$\frac{R_2^2}{4h} = 2h$$
$$R_2^2 = 8h^2$$

Taking the square root gives us the new maximum radius:
$$R_2 = 2\sqrt{2}h$$

### The Final Area Comparison

With the new maximum range $R_2$, we can calculate the new watered area $A_2$. Again, the water forms a circle on the ground:
$$A_2 = \pi R_2^2$$

Substituting $R_2^2 = 8h^2$:
$$A_2 = \pi (8h^2) = 8\pi h^2$$

Finally, we want to find the factor by which the watered area has changed. We simply divide the new area by the initial area:
$$	ext{Factor} = \frac{A_2}{A_1} = \frac{8\pi h^2}{4\pi h^2} = 2$$

The watered area exactly doubles when the sprinkler is raised to a height equal to its maximum vertical reach. The final answer is **2**.

Enter Numerical Value:

Visualized Solution

$v = 2 g h$

$A_{1} = 4 π h^{2}$

$y = h + \frac{v ^{2}}{2 g} - \frac{g x ^{2}}{2 v ^{2}}$

$R_{2} = 22 h$

$\frac{A _{2}}{A _{1}} = 2$

The Sigma Insight: Projectile Motion

Visualizing the Sprinkler's Reach

The Ground Level Setup

Elevating the Sprinkler

The Envelope of Trajectories

The Final Area Comparison

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Enter Numerical Value:

Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​=4πh2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=h+2gv2​−2v2gx2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​=22​h

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​A2​​=2