A fountain jet situated at a height of 11.25 m above the ground projects water droplets in all directions with velocity of 20 m/s. Calculate the maximum distance (in m) on ground reached by any water droplet

Options

A. 50 m
B. 100 m
C. 75 m
D. 25 m

Explanation

To solve the problem of calculating the maximum distance on the ground reached by any water droplet projected from a fountain jet situated at a height of 11.25 m with an initial velocity of 20 m/s, we can follow these steps:

### Step 1: Understand the Problem
The fountain is at a height of 11.25 m and projects water droplets at a speed of 20 m/s in all directions. We need to find the maximum horizontal distance (range) that any droplet can reach when projected at an angle.

### Step 2: Break Down the Motion
The motion of the water droplet can be analyzed in two components:
1. Vertical motion (affected by gravity)
2. Horizontal motion (constant velocity)

### Step 3: Determine the Time of Flight
The vertical motion can be described using the kinematic equation:
\[ y = v_{y} t - \frac{1}{2} g t^2 \]
Where:
- \( y \) is the vertical displacement (which will be -11.25 m, as it falls down)
- \( v_{y} \) is the initial vertical velocity component, which can be expressed as \( v \sin \theta \)
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( t \) is the time of flight

Setting up the equation:
\[ -11.25 = (20 \sin \theta) t - \frac{1}{2} (9.81) t^2 \]

### Step 4: Solve for Time
Rearranging gives us a quadratic equation in terms of \( t \):
\[ \frac{1}{2} (9.81) t^2 - (20 \sin \theta) t - 11.25 = 0 \]

Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- \( a = \frac{1}{2} (9.81) \)
- \( b = -20 \sin \theta \)
- \( c = -11.25 \)

### Step 5: Find the Horizontal Distance
The horizontal distance \( x \) can be expressed as:
\[ x = v_{x} t \]
Where \( v_{x} = v \cos \theta \) and \( v = 20 \, \text{m/s} \).

### Step 6: Maximize the Range
To find the maximum distance, we need to maximize \( x \). The maximum range occurs when the angle \( \theta \) is 45 degrees (for projectile motion). Thus, we can substitute \( \sin(45^\circ) = \cos(45^\circ) = \frac{1}{\sqrt{2}} \).

### Step 7: Calculate Time of Flight for \( \theta = 45^\circ \)
Substituting \( \sin(45^\circ) \) into the time of flight equation:
\[ -11.25 = (20 \cdot \frac{1}{\sqrt{2}}) t - \frac{1}{2} (9.81) t^2 \]

### Step 8: Solve for Time
This will yield a specific time \( t \) that can be substituted back into the horizontal distance formula.

### Step 9: Calculate the Maximum Distance
Substituting the time back into the horizontal distance equation will yield the maximum distance \( x \).

### Final Calculation
After solving the equations, we find that the maximum horizontal distance \( x \) is approximately 50 m.

### Summary
The maximum distance on the ground reached by any water droplet is **50 m**.

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