A water fountain on the ground sprinkles water all around it. If the speed of water coming out of th

Q: Why is the wet area circular?

Because the fountain shoots water symmetrically in all directions with the same initial speed. The maximum distance reached in any direction will be the same, forming a circle.

Q: Why do we take the angle as 45 degrees?

To find the *total* area that gets wet, we need the outermost boundary, which corresponds to the maximum possible range. For a given speed, the range is maximum when the launch angle is 45 degrees.

Q: Does the height of the nozzle matter?

In this specific problem, the fountain is stated to be 'on the ground', so we assume the nozzle height is negligible. If it had a significant height, the maximum range formula would be different.

Q: What if the water speed $v$ varies?

The problem states a constant speed $v$. If it varied, the maximum speed would determine the outermost boundary of the wet area.

Q: Why isn't the area just $\pi v^2 / g$?

You forgot to square the radius! The radius itself is $v^2/g$. When you plug it into the area formula $A = \pi r^2$, you must square the entire term, which gives $v^4/g^2$.

Question

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is $v$, the total area around the fountain that gets wet is

Options:

(A) $\pi \frac{v^4}{g^2}$

(B) $\frac{\pi}{2} \frac{v^4}{g^2}$

(C) $\pi \frac{v^2}{g^2}$

(D) $\pi \frac{v^2}{g}$

Sigma Percentile · Accepted Answer

Have you ever walked past a beautiful water fountain in a park and wondered about the physics governing its elegant display? The way water droplets arc through the air and splash onto the ground is a perfect real-world demonstration of projectile motion.

In this problem, we are asked to find the total area around a ground-level fountain that gets wet. The fountain shoots water in all directions with a constant initial speed $v$.

### Visualizing the Fountain

Imagine standing right next to the fountain. Water is being sprayed outwards in a full $360^\circ$ circle. Every single droplet of water acts as a tiny projectile, launched with an initial velocity $v$ at some angle $	heta$ with respect to the horizontal.

Because the water is sprayed symmetrically in all directions, the wet patch on the ground will form a perfect circle. To find the area of this circle, we need to determine its radius. The radius of this wet area is simply the maximum horizontal distance that any water droplet can travel.

### The Quest for Maximum Range

In kinematics, the horizontal distance traveled by a projectile is called its range. The formula for the range $R$ of a projectile launched from the ground is given by:

$$R = \frac{v^2 \sin(2	heta)}{g}$$

Here, $v$ is the launch speed, $	heta$ is the launch angle, and $g$ is the acceleration due to gravity.

To maximize the wet area, we need to maximize this range. Looking at the formula, the range is maximum when the term $\sin(2	heta)$ is at its maximum value. The maximum value of the sine function is $1$, which occurs when the angle is $90^\circ$.

Therefore, we set $2	heta = 90^\circ$, which gives us a launch angle of $	heta = 45^\circ$.

Substituting this back into our range formula, we find the maximum possible radius for our wet circle:

$$R_{\max} = \frac{v^2 \sin(90^\circ)}{g} = \frac{v^2}{g}$$

### Calculating the Wet Area

Now that we have the radius of the circular wet patch, finding the total area is straightforward geometry. The area $A$ of a circle is given by $\pi$ times the square of its radius:

$$A = \pi R_{\max}^2$$

We substitute our expression for $R_{\max}$ into this area formula:

$$A = \pi \left(\frac{v^2}{g}ight)^2$$

Squaring the terms inside the parenthesis, we arrive at our final, elegant result:

**$$A = \pi \frac{v^4}{g^2}$$**

This tells us that the wet area is highly sensitive to the speed of the water. If you double the speed of the fountain, the wet area doesn't just double—it increases by a factor of sixteen!

Animated Solution for Physics - Kinematics: A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is $v$ , the total area around the fountain that gets wet is

Select Answer:

Visualized Solution

Visualizing the Fountain

Identifying the Boundary

Condition for Maximum Range

Calculating Maximum Range

Area of the Wet Patch

Final Substitution

Food for Thought

The Sigma Insight: Projectile Motion

Visualizing the Fountain

The Quest for Maximum Range

Calculating the Wet Area

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Select Answer:

Visualized Solution

Visualizing the Fountain

Identifying the Boundary

Condition for Maximum Range

Calculating Maximum Range

Area of the Wet Patch

Final Substitution

Food for Thought

The Sigma Insight: Projectile Motion

Visualizing the Fountain

The Quest for Maximum Range

Calculating the Wet Area

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