Question

What is the distance traveled by a water jet from a sprinkler mounted on the ground, which is sent out at a 20° angle to the horizontal with a speed of 11 m/s?

Answer 1

To find the distance traveled by the water jet from the sprinkler, we can use the physics of projectile motion.

1. **Break down the initial velocity into horizontal and vertical components**:
- The horizontal component of the velocity (\(v_{0x}\)):
\[
v_{0x} = v_0 \cdot \cos(θ) = 11 \, \text{m/s} \cdot \cos(20°)
\]
- The vertical component of the velocity (\(v_{0y}\)):
\[
v_{0y} = v_0 \cdot \sin(θ) = 11 \, \text{m/s} \cdot \sin(20°)
\]

2. **Calculate these components**:
- Using a calculator:
\[
v_{0x} = 11 \cdot \cos(20°) \approx 11 \cdot 0.9397 \approx 10.34 \, \text{m/s}
\]
\[
v_{0y} = 11 \cdot \sin(20°) \approx 11 \cdot 0.3420 \approx 3.76 \, \text{m/s}
\]

3. **Calculate the time of flight**:
The time of flight \(t\) for a projectile launched and landing at the same height can be calculated using:
\[
t = \frac{2 \cdot v_{0y}}{g}
\]
where \(g \approx 9.81 \, \text{m/s}^2\) is the acceleration due to gravity.
\[
t = \frac{2 \cdot 3.76}{9.81} \approx \frac{7.52}{9.81} \approx 0.766 \, \text{s}
\]

4. **Calculate the horizontal distance traveled (range)**:
The horizontal distance \(R\) can be found using:
\[
R = v_{0x} \cdot t
\]
Substituting the values:
\[
R = 10.34 \, \text{m/s} \cdot 0.766 \, \text{s} \approx 7.91 \, \text{m}
\]

Therefore, the distance traveled by the water jet from the sprinkler is approximately **7.91 meters**.