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(III) Assume a net force \(F=-m g-k v^{2}\) acts during the upward vertical motion of a \(250-\mathrm{kg}\) rocket, starting at the moment \((t=0)\) when the fuel has burned out and the rocket has an upward speed of \(120 \mathrm{~m} / \mathrm{s}\). Let \(k=0.65 \mathrm{~kg} / \mathrm{m}\). Estimate \(v\) and \(y\) at 1.0 -s intervals for the upward motion only, and estimate the maximum height reached. Compare to free-flight conditions without air resistance \((k=0)\).

Short Answer

Expert verified
With air resistance, max height is less. Without, calculate using energy conservation.

Step by step solution

01

Understand the forces at play

We have two forces acting on the rocket: gravitational force given by \(mg\) and air resistance proportional to \(v^2\) given by \(kv^2\). These combine to form the net force formula: \[ F = -mg - kv^2 \]where \(m = 250\) kg, \(g = 9.8\) m/s², and \(k = 0.65\) kg/m.
02

Set up the equations of motion

The acceleration \(a\) is the net force divided by the mass, \[ a = \frac{F}{m} = -g - \frac{k}{m}v^2 \]Thus, \[ a = -9.8 - \frac{0.65}{250}v^2 \] mm/s². This equation will help us determine velocity changes over time.
03

Estimate velocity at different intervals

Start with initial velocity \(v_0 = 120\) m/s at \(t = 0\). For each second up to the maximum height, update velocity using \[ v = v_0 + a \cdot \Delta t \]for each 1 s iteration. Adjust \(v\) using the current value of \(a\) calculated with the current \(v\) value.
04

Calculate position at different intervals

Use the average velocity over each interval to estimate position. For each time interval:\[ y = y_0 + v_{avg} \cdot \Delta t \]where \(v_{avg}\) is the mean of initial and final velocity for the interval: \[ v_{avg} = \frac{v_0 + v}{2} \]. Update \(y_0\) to the new position each time.
05

Determine the maximum height

Continue iterating until the velocity falls to zero, indicating the rocket has stopped moving upward. This final position value is the maximum height achieved with air resistance.For **no air resistance** (\(k=0\)), use:\[ v(t) = v_0 - gt \]and the maximum height can be calculated when \(v(t) = 0\):\[ t = \frac{v_0}{g} \].The height is \[ y = v_0 \cdot \frac{v_0}{g} - \frac{1}{2}g \left(\frac{v_0}{g}\right)^2 \].

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Key Concepts

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

Rocket Motion
Rocket motion involves the study of objects moving mainly due to propulsion forces, which often occur in a vacuum or atmosphere. When a rocket flies upwards after burning its fuel, it continues on its trajectory under the action of various forces.
These forces include gravity and air resistance, which impact its motion through complex interactions.
Understanding rocket motion requires using physics principles to compute changes in velocity and height as the rocket progresses through its ascent. The initial conditions, such as starting velocity and mass, play key roles in predicting the rocket's upward journey.
  • During rocket motion, the absence of propulsion means external forces gradually slow the rocket.
  • Gravity consistently acts downward, pulling the rocket back to Earth.
  • Air resistance varies with speed and acts against the direction of motion, reducing velocity over time.
Equations of Motion
The equations of motion are fundamental rules that describe how objects move based on initial parameters and external forces. They enable us to predict future states of motion like velocity and position.
In the case of the rocket, we use specific equations to account for gravity and air resistance.
When analyzing the rocket, the key equation used is
\[ a = -g - \frac{k}{m}v^2 \]
where:
  • \(a\) is acceleration.
  • \(g\) represents the gravitational acceleration (9.8 m/s²).
  • \(k\) is the air resistance coefficient.
  • \(m\) denotes the rocket's mass.
This differential equation helps trace the rocket's deceleration path until it ceases upward motion. By integrating this equation over time intervals, we can update the rocket's velocity and height.
These calculations increase in complexity when incorporating air resistance, needing iterative steps to achieve accuracy.
Air Resistance
Air resistance, also known as drag, represents the forces that objects encounter as they move through the atmosphere. This resistance rapidly impacts objects like rockets and often depends on their speed and shape.
For a rocket moving vertically, the air resistance is proportional to the square of its speed, given by the term \(kv^2\). Here, \(k\) plays a crucial role, defining how strongly the atmosphere resists the rocket's motion.
  • Air resistance acts in the opposite direction to the rocket's motion, decelerating it.
  • As the rocket rises and decelerates, air resistance gradually decreases.
  • Adjustments in speed calculations require incorporating air resistance to predict exact position and speed over time accurately.
Integrating air resistance effectively requires modifying the motion equations to reflect its impact, presenting a more realistic motion model.
Gravitational Force
Gravitational force is the natural phenomenon by which objects with mass are attracted toward each other. On Earth, this force pulls everything directly toward its center.
In rocket motion, gravity plays a pivotal role as it continually influences the rocket's path by applying a downward force.
For our specific problem, the gravitational force is expressed by \(mg\), where \(m\) is the mass of the rocket (250 kg) and \(g\) is the acceleration due to Earth's gravity (9.8 m/s²).
Understanding gravitational force is crucial for calculating how far and how fast a rocket can travel.
  • It continuously acts on the rocket, causing it to slow as it ascends.
  • Gravity’s influence means the rocket must overcome this force to achieve any upward velocity or height.
  • Gravitational calculations are integral in determining maximum height and trajectory.
Understanding this force allows scientists and engineers to effectively design rockets to reach desired altitudes, accounting for this natural resistance to upward movement.

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Most popular questions from this chapter

(II) Two crates, of mass 65 \(\mathrm{kg}\) and 125 \(\mathrm{kg}\) , are in contact and at rest on a horizontal surface (Fig, \(32 ) . \mathrm{A} 650\) -N force is exerted on the 65 -kg crate. If the coefficient of kinetic friction is \(0.18,\) calculate \((a)\) the acceleration of the system, and \((b)\) the force that each crate exerts on the other. (c) Repeat with the crates reversed.

A car drives at a constant speed around a banked circular track with a diameter of 127 \(\mathrm{m}\) . The motion of the car can be described in a coordinate system with its origin at the center of the circle. At a particular instant the car's accel- eration in the horizontal plane is given by $$\vec{\mathbf{a}}=(-15.7 \hat{\mathbf{i}}-23.2 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2}$$ (a) What is the car's speed? (b) Where \((x\) and \(y)\) is the car at this instant?

A \(72-\mathrm{kg}\) water skier is being accelerated by a ski boat on a flat ("glassy") lake. The coefficient of kinetic friction between the skier's skis and the water surface is \(\mu_{\mathrm{k}}=0.25\) (Fig. \(5-55)\). (a) What is the skier's acceleration if the rope pulling the skier behind the boat applies a horizontal tension force of magnitude \(F_{\mathrm{T}}=240 \mathrm{~N}\) to the skier \(\left(\theta=0^{\circ}\right) ?\) (b) What is the skier's horizontal acceleration if the rope pulling the skier exerts a force of \(F_{\mathrm{T}}=240 \mathrm{~N}\) on the skier at an upward angle \(\theta=12^{\circ} ?\) (c) Explain why the skier's acceleration in part ( \(b\) ) is greater than that in part ( \(a\) ).

(1I) Police investigators, examining the scene of an accident involving two cars, measure 72 -m-long skid marks of one of the cars, which nearly came to a stop before colliding. The coefficient of kinetic friction between rubber and the pave- ment is about \(0.80 .\) Estimate the initial speed of that car assuming a level road.

(II) Use dimensional analysis (Section \(1-7\) ) to obtain the form for the centripetal acceleration, \(a_{\mathrm{R}}=v^{2} / r\)

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