Calculating the Height at Which a Stream of Water Strikes a Wall: An Analogue Physics Example

In physics, understanding the motion of projectiles is an essential concept. This article delves into a practical example involving a stream of water shooting upward at a specific angle and speed. We will demonstrate the step-by-step process of calculating the height at which the water will strike a wall. Through this example, we will explore the fundamental principles of projectile motion and how to separate velocity components to solve complex physics problems.

Understanding the Problem

The scenario involves a hose lying on the ground, shooting a stream of water upward at a 40° angle to the horizon. The initial speed of the water as it leaves the hose is 20 m/s. The challenge is to determine how high the stream of water will strike a wall that is 8 meters away horizontally. This problem can be broken down into two main components: horizontal and vertical motion.

Step-by-Step Solution

To solve the problem, we need to follow several steps, including decomposing the initial velocity, calculating the time to reach the wall, and finally determining the height at which the water strikes the wall.

Step 1: Decomposing the Initial Velocity

The initial velocity of the water is 20 m/s at an angle of 40°. We can separate this velocity into horizontal and vertical components using trigonometry.

Horizontal Component:

v_x v_0 cdot cos(40°) 20 cdot cos(40°) approx 15.32 text{ m/s}

Vertical Component:

v_{0y} v_0 cdot sin(40°) 20 cdot sin(40°) approx 12.86 text{ m/s}

Step 2: Calculating the Time to Reach the Wall

The horizontal distance to the wall is 8 meters. To calculate the time it takes for the water to reach the wall, we use the horizontal motion equation.

t frac{text{horizontal distance}}{v_x} frac{8}{15.32} approx 0.521 text{ s}

Step 3: Calculating the Height at That Time

To find the height at which the water strikes the wall, we use the vertical motion equation, considering the acceleration due to gravity.

y v_{0y} cdot t - frac{1}{2} g t^2

Substituting the values, we get:

First Term:

12.86 cdot 0.521 approx 6.69 text{ m}

Second Term:

frac{1}{2} cdot 9.81 cdot 0.521^2 approx 1.33 text{ m}

Therefore, the height at which the water strikes the wall is:

y approx 6.69 - 1.33 approx 5.36 text{ m}

Conclusion

The water will strike the wall at approximately 5.36 meters. This result is crucial for understanding the behavior of projectiles and the importance of separating velocity components in physics problems.

Advanced Calculations for Physics Enthusiasts

For those interested in a more detailed analysis, the parameters of the water when it strikes the wall can be further dissected:

Time [t]

t frac{text{range}}{v cos(theta)} frac{8}{20 cos(40°)} 0.522163 text{ s}

Height [h]

h v_{0y} t - frac{1}{2} g t^2 20 sin(40°) cdot 0.522163 - frac{1}{2} cdot 9.80665 cdot 0.522163^2 approx 5.375886 text{ m}

Horizontal Velocity [Vx]

V_x v cos(theta) 20 cos(40°) approx 15.320889 text{ m/s}

Vertical Velocity [Vy]

V_y v_{0y} - g t 20 sin(40°) - 9.80665 cdot 0.522163 approx 7.735083 text{ m/s}

Resultant Velocity

text{Resultant velocity} sqrt{V_x^2 V_y^2} sqrt{15.320889^2 7.735083^2} approx 17.162784 text{ m/s}

Final Velocity Angle

text{Angle} arctanleft(frac{V_y}{V_x}right) arctanleft(frac{7.735083}{15.320889}right) approx 26.787918°

This comprehensive analysis provides a deeper understanding of the physics involved and can be applied to various real-world scenarios.