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A rocket is launched vertically from rest with a constant upward acceleration of 1.75 \(\mathrm{m} / \mathrm{s}^{2} .\) Suddenly 22.0 \(\mathrm{s}\) after launch, an unneeded fuel tank is jettisoned by shooting it away from the rocket. Acrew member riding in the rocket measures that the initial speed of the tank is 25.0 \(\mathrm{m} / \mathrm{s}\) and that it moves perpendicular to the rocket's path. The fuel tank feels no appreciable air resistance and reels only the force of gravity once it leaves the rocket. (a) How feels only the force of gravity once it leaves the rocket. (a) How fast is the rocket moving at the instant the fuel tank is jettisoned? (b) What are the horizontal and vertical componed by (i) a crew unk's velocity just as it is jetisoned as measured by (i) a crew member in the rocket and (ii) a technician standing on the ground? (c) At what angle with respect to the horizontal does the jettisoned fuel tank initially move, as measured by a crew member in the rocket and (ii) a technician standing on the ground? (d) What maximum height above the launch pad does the jettisoned tank reach?

Step by step solution

01

Calculate the Rocket's Velocity

To find the rocket's velocity when the fuel tank is jettisoned, use the equation for velocity in uniformly accelerated motion: \( v = u + at \), where \( u = 0 \ \mathrm{m/s} \) (since it starts from rest), \( a = 1.75 \ \mathrm{m/s^2} \), and \( t = 22.0 \ \mathrm{s} \). Plug these values into the equation: \[ v = 0 + (1.75 \, \mathrm{m/s^2}) \times 22.0 \, \mathrm{s} = 38.5 \, \mathrm{m/s} \] Thus, the rocket is moving at 38.5 \( \mathrm{m/s} \) at the instant the fuel tank is jettisoned.

02

Determine Velocity Components of the Tank (Relative to Rocket)

Relative to the rocket, the tank is ejected perpendicularly with an initial speed of 25.0 \( \mathrm{m/s} \). Since this speed is perpendicular to the motion of the rocket, it only affects the horizontal component of the tank's velocity: \( v_{x} = 25.0 \, \mathrm{m/s} \), and \( v_{y} = 38.5 \, \mathrm{m/s} \) (same as the rocket).

03

Calculate Tank's Velocity Components (Relative to Ground)

To find the tank's velocity relative to the ground, we combine its horizontal and vertical components. The horizontal component is \( v_{x} = 25.0 \, \mathrm{m/s} \), and the vertical component is the sum of the vertical velocity of the rocket and the perpendicular velocity: \( v_{y} = 38.5 \, \mathrm{m/s} \).

04

Find the Angle of the Tank's Initial Motion (Crew's Perspective)

For the crew in the rocket, the angle \( \theta \) of the tank's launch can be found using \( \tan^{-1} \left( \frac{v_{y}}{v_{x}} \right) \). Here, \( v_{y} = 38.5 \, \mathrm{m/s} \) and \( v_{x} = 0 \, \mathrm{m/s} \): \[ \theta = \tan^{-1} \left(\frac{38.5}{25}\right) \approx 56.31^\circ \]

05

Find the Angle of the Tank's Initial Motion (Ground Perspective)

For a technician on the ground, calculate \( \theta \) using: \( \tan^{-1} \left( \frac{38.5}{25} \right)\). Again, this results in approximately \( 56.31^\circ \).

06

Calculate Maximum Height Reached by the Tank

The maximum additional height \( h \) attained by the tank can be found using the kinematic equation \( v_{y}^2 = u_{y}^2 + 2ah \), where the final vertical velocity \( v_{y} = 0 \ \mathrm{m/s} \) at peak height, \( u_{y} = 38.5 \, \mathrm{m/s} \), and \( a = -9.8 \, \mathrm{m/s^2} \). Solving for \( h \): \[ 0 = (38.5)^2 + 2(-9.8)h \] \[ h = \frac{(38.5)^2}{2 \times 9.8} \approx 75.6 \, \mathrm{m} \]

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Key Concepts

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

Kinematic Equations

Kinematic equations are the backbone of physics problems involving motion. In this exercise, we used the equation that connects initial velocity (\( u \)), acceleration (\( a \)), time (\( t \)), and final velocity (\( v \)): \[ v = u + at \]Understanding this equation is crucial. It starts from the concept of changing velocity due to acceleration over time. Since the rocket initiates from rest, the initial velocity is zero. Therefore, the velocity after 22 seconds is simply the product of acceleration and time. This elegant simplicity allows us to predict the rocket’s speed at any time.

Velocity Components

In projectile motion, determining velocity components is key to solving the problem. Every velocity vector has horizontal and vertical components. For the fuel tank ejection, this concept is very useful. When the fuel tank is jettisoned, it has both horizontal and vertical velocities. Calculating these components requires understanding the relative motion.

• Horizontally, the tank moves at 25 \( \text{m/s} \), as it shoots away perpendicularly.
• Vertically, it matches the rocket’s speed of 38.5 \( \text{m/s} \) because of its shared journey until the ejection.

These components are vital for understanding how the tank moves through space both immediately after ejection and throughout its trajectory.

Vertical and Horizontal Motion

Projectile motion separates into vertical and horizontal components, each following different rules. The horizontal motion of the tank remains unaffected by gravity initially, as it moves horizontally at a constant speed of 25 \( \text{m/s} \). Vertical motion, however, is governed by gravity. When you separate the problem into these dimensions:

• Horizontal motion is simple: there's no acceleration in the absence of air resistance, keeping speed constant.
• Vertical motion mirrors free-fall problems, where acceleration due to gravity acts downwards.

This separation simplifies problems into more manageable pieces and helps in applying the correct equations to each component.

Uniformly Accelerated Motion

Uniformly accelerated motion describes a scenario where acceleration is constant. Think about our rocket’s departure: it accelerates upwards at 1.75 \( \text{m/s}^2 \). This steady acceleration defines the rocket's upward motion, allowing us to predict its velocity over time using kinematic equations. However, the ejected fuel tank experiences uniformly accelerated motion in another way:

• After separation, it undergoes uniformly accelerated motion downward due to gravity (-9.8 \( \text{m/s}^2 \)).

Such scenarios highlight uniform acceleration’s predictability in dynamics, essential for calculating future positions or velocities of moving objects.