You hold a hose 45 degrees to the horizontal and at a height of 1 m from the floor. The water reaches a maximum distance of 10 m from where you are standing. Now you place your thumb over the end of the hose to occlude the opening by 80%, which in turn reduces the flow rate by 50%. Even though less fluid emerges the water travels further. How far does the water travel with your thumb over the end of the hose? Find the amount by which the hose must be occluded in order for the water to travel twice as far.
To answer these questions, we need to understand the principles of fluid dynamics and how they apply to the given scenario.
First, let's address the initial situation where the hose is held at a 45-degree angle to the horizontal and a height of 1 m from the floor. The water reaches a maximum distance of 10 m from where you are standing. We can use some basic trigonometry to calculate the initial horizontal velocity of the water.
The horizontal distance traveled by the water can be represented by the equation:
distance = velocity × time
Since we want to find the distance, we rearrange the equation:
velocity = distance / time
We know that the distance traveled is 10 m, but we need to determine the time it takes for the water to reach that distance. To do that, we need to analyze the motion of the water.
When the water leaves the hose, it follows a parabolic trajectory known as a projectile motion. In this case, we can assume that the water is only affected by gravity, neglecting air resistance.
The horizontal velocity of the water remains constant throughout its path, unaffected by gravity. This is because there are no forces acting horizontally. The vertical velocity of the water, however, changes due to the force of gravity.
At the peak of the water's trajectory, the vertical velocity becomes zero. We can calculate the time it takes to reach the peak by considering the vertical motion alone:
0 = initial vertical velocity + (acceleration due to gravity) × time
The initial vertical velocity is zero (as the water is not initially moving vertically), and the acceleration due to gravity is -9.8 m/s² (assuming a downward direction). Solving this equation gives us the time it takes for the water to reach its peak.
Now, we can double this time, as the water will take the same amount of time to reach the maximum distance on the way down. This gives us the total time of flight.
Using this total time of flight, we can calculate the initial horizontal velocity of the water using the equation mentioned earlier.
Now, let's move on to the second situation where your thumb occludes the opening of the hose by 80% and reduces the flow rate by 50%.
When the opening of the hose is occluded, less fluid can pass through, reducing the flow rate. The flow rate is directly proportional to the cross-sectional area of the hose opening. Assuming the hose has a circular cross-section, the flow rate is given by:
flow rate = cross-sectional area × velocity
Now, when your thumb occludes the opening by 80%, the new flow rate will be reduced by 50%. We can represent this as:
new flow rate = (original flow rate) × (1 - 0.8) × (1 - 0.5)
To find the new distance traveled, we need to calculate the new velocity. Rearranging the equation above, we have:
new velocity = (new flow rate) / cross-sectional area
To find the amount by which the hose must be occluded in order for the water to travel twice as far, we can set up a proportion:
(distance with thumb occlusion) / 10 = 2 / 1
Solving this proportion will give us the desired amount of occlusion.
By following these steps and using the relevant equations and principles, you should be able to find the answers to your questions.