/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 A jet of water leaving a nozzle ... [FREE SOLUTION] | 91Ó°ÊÓ

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A jet of water leaving a nozzle is directed vertically upward. The jet leaves the nozzle with a speed of \(30 \mathrm{ft} / \mathrm{s}\). Neglect thermal and friction effects. How far above the nozzle will the water travel?

Short Answer

Expert verified
The water reaches a maximum height of 14.0625 ft above the nozzle.

Step by step solution

01

Identify the Known Variables

We are given that the initial velocity of the jet of water is \(v_0 = 30\, \text{ft/s}\). The acceleration due to gravity is \(g = -32\, \text{ft/s}^2\) because it acts downward against the upward motion of the jet. The final velocity at the highest point will be \(v = 0\, \text{ft/s}\).
02

Use Kinematic Equation

We use the kinematic equation \(v^2 = v_0^2 + 2a s\), where \(s\) is the distance (height in this case) traveled, \(a\) is acceleration, and \(v\) is the final velocity. Our aim is to solve for \(s\).
03

Substitute the Known Values

Substitute the values into the kinematic equation: \((0)^2 = (30)^2 + 2(-32) s\).
04

Solve for the Maximum Height

Simplify the equation: \(0 = 900 - 64s\). Solve for \(s\) by rearranging the equation to find \(s\): \(64s = 900\), which gives \(s = \frac{900}{64} = 14.0625\).
05

Conclusion

The maximum height reached by the water above the nozzle is \(14.0625\, \text{ft}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Projectile Motion
Projectile motion refers to the motion of an object that is thrown or projected into the air, subject to only the acceleration of gravity. Importantly, projectile motion involves two components of flight: horizontal and vertical. In the context of our problem, the water jet represents a simplified form of projectile motion where we focus mainly on the vertical component because the water is traveling straight up and down.

When dealing with projectile motion, we often use kinematic equations to determine unknown variables such as time of flight, maximum height, or range. Remember that in ideal projectile motion problems like this one, we neglect air resistance and other forces except for gravity.
Vertical Motion
Vertical motion is a key component of projectile motion where an object moves up and down under the influence of gravity. When the water jet is directed vertically upwards from the nozzle, it follows a path determined by its initial velocity and acceleration due to gravity.

As the water moves upward, its speed decreases due to the pull of gravity acting downwards. This deceleration continues until it reaches the highest point, known as the apex, where its velocity temporarily becomes zero before gravity pulls it back down.

In solving for vertical motion problems, we often start by identifying known variables such as initial velocity and acceleration. We then apply specific kinematic equations to determine unknowns such as the maximum height in this case.
Gravity
Gravity is a constant force acting on objects near the Earth's surface, pulling them towards the center of the Earth. In many physics problems, gravity is considered to act uniformly across the object’s trajectory. It is this consistent force that affects the motion of all projectiles, including the upward journey of our water jet.

The acceleration due to gravity is commonly denoted as 'g' and has a value of approximately \(-9.8\, \text{m/s}^2\) in the metric system or \(-32\, \text{ft/s}^2\) in the imperial system, as used in our exercise. This negative sign indicates the downward direction of the force. Understanding gravity’s role allows us to predict and calculate changes in the object's velocity and position over time.
Kinematic Equations
Kinematic equations are essential tools in solving motion problems, particularly those involving constant acceleration, such as projectile motion. They allow us to relate different variables: initial velocity, final velocity, acceleration, time, and displacement.

In our scenario, we used the kinematic equation:\[v^2 = v_0^2 + 2a s\]where:
  • \(v\) is the final velocity, which is zero at the highest point.
  • \(v_0\) is the initial velocity, given as \(30\, \text{ft/s}\).
  • \(a\) is the acceleration, here it's the gravity \(-32\, \text{ft/s}^2\).
  • \(s\) is the displacement or height reached by the object.
By substituting the known values, we solved for \(s\) or the maximum height the water reached. These equations are invaluable in predicting and describing the behavior of moving objects in a variety of circumstances.

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Most popular questions from this chapter

An isentropic gas moves through a horizontal pipe of constant cross-sectional area. If the inlet pressure is \(12 \mathrm{psi}\) with a velocity of \(70 \mathrm{ft} / \mathrm{s}\), and the exit pressure is \(1 \mathrm{psi}\), calculate the exit velocity of gas, given the ratio of specific heats \(k=1.65\)

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