Question

Calculate the maximum height to which water could be squirted with the hose if it emerges from the nozzle at 16.2 m/s.

Answer 1

The maximum height to which water could be squirted from the hose, if it emerges from the nozzle at 16.2 m/s, is calculated using the conservation of energy principle and is found to be approximately 13.4 meters.

Step-by-step explanation:

To calculate the maximum height to which water could be squirted with the hose if it emerges from the nozzle at a speed of 16.2 m/s, we can use the physics principle of conservation of energy. The water starts with kinetic energy and will convert this energy into potential energy at the highest point of its trajectory.

The initial kinetic energy (KE) of the water can be given by the equation KE = (1/2)mv², where m is the mass of the water and v is the velocity at which it emerges from the nozzle. Upon reaching its maximum height h, the water will have potential energy (PE) equal to mgh where g is the acceleration due to gravity (9.81 m/s²). Setting the initial kinetic energy equal to the potential energy at the maximum height gives us: (1/2)mv² = mgh. Solving for h, we can cancel out m from both sides and get: h = v² / (2g).

Plugging in the value of v = 16.2 m/s, we get h = (16.2 m/s)² / (2*9.81 m/s²) = 13.4 meters. This is the maximum height to which the water could be squirted from the nozzle.

Answer 2

The maximum height to which water could be squirted with the hose is 13.380 meters.

Step-by-step explanation:

A line of current of a fluid can be explained sufficiently by Bernoulli's Theorem. In this case, the system can be simplified due to neglectance of changes in absolute pressure. Water is squirted with an initial speed and reaches its maximum height, where final speed is zero. That is to say:

`(v_(1)^(2))/(2\cdot g) +z_(1) = (v_(2)^(2))/(2\cdot g) +z_(2)`

Where:

`z_(1)`,
`z_(2)` - Initial and final height of water, measured in meters.

`g` - Gravitational acceleration, measured in meters per square second.

`v_(1)`,
`v_(2)` - Initial and final speed of water, measured in meters per second.

If
`z_(1) = 0\,m`,
`v_(1) = 16.2\,(m)/(s)`,
`v_(2) = 0\,(m)/(s)` and
`g = 9.807\,(m)/(s^(2))`, then:

`z_(2) = (v_(1)^(2)-v_(2)^(2))/(2\cdot g) +z_(1)`

`z_(2) = (\left(16.2\,(m)/(s) \right)^(2)-\left(0\,(m)/(s) \right)^(2))/(2\cdot \left(9.807\,(m)/(s^(2)) \right)) +0\,m`

`z_(2) = 13.380\,m`

The maximum height to which water could be squirted with the hose is 13.380 meters.