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When it is 145 \(\mathrm{m}\) above the ground, a rocket traveling vertically upward at a constant 8.50 \(\mathrm{m} / \mathrm{s}\) relative to the ground launches a secondary rocket at a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(53.0^{\circ}\) above the horizontal, both quantities being measured by an astronaut sitting in the rocket. After it is launched the secondary rocket is in free-fall. (a) Just as the secondary rocket is launched, what are the horizontal and vertical components of its velocity relative to (i) the astronaut sitting in the rocket and (ii) Mission Control on the ground? (b) Find the initial speed and launch angle of the secondary rocket as measured by Mission Control. (c) What maximum height above the ground does the secondary rocket reach?

Step by step solution

01

Calculate horizontal and vertical components of velocity relative to astronaut

To find the components of the secondary rocket's velocity relative to the astronaut, we use the given speed of 12.0 m/s and the angle of 53.0°. The horizontal component (vx_astronaut) is given by \( v_{x_{ ext{astronaut}}} = 12.0 \cos(53.0°) \approx 7.21 \text{ m/s} \) and the vertical component (vy_astronaut) is given by \( v_{y_{ ext{astronaut}}} = 12.0 \sin(53.0°) \approx 9.60 \text{ m/s} \).

02

Calculate velocity components relative to Mission Control for horizontal motion

Since the rocket from where the secondary rocket is launched is moving vertically, the horizontal component of the secondary rocket relative to the astronaut equals the relative component to Mission Control. Thus, for horizontal motion, \( v_{x_{ ext{ground}}} = v_{x_{ ext{astronaut}}} = 7.21 \text{ m/s} \).

03

Calculate velocity components relative to Mission Control for vertical motion

The vertical velocity relative to Mission Control involves adding the velocity of the main rocket. Thus, the vertical velocity component is \( v_{y_{ ext{ground}}} = v_{y_{ ext{astronaut}}} + 8.50 = 9.60 + 8.50 = 18.10 \text{ m/s} \).

04

Calculate the initial speed relative to Mission Control

The initial speed, \( v_i \), relative to the ground is found using the Pythagorean theorem: \( v_i = \sqrt{v_{x_{ ext{ground}}}^2 + v_{y_{ ext{ground}}}^2} \approx \sqrt{(7.21)^2 + (18.10)^2} \approx 19.45 \text{ m/s} \).

05

Calculate launch angle relative to Mission Control

The launch angle, \( \theta \), from the horizontal relative to Mission Control is \( \theta = \tan^{-1}\left(\frac{v_{y_{ ext{ground}}}}{v_{x_{ ext{ground}}}}\right) \approx \tan^{-1}\left(\frac{18.10}{7.21}\right) \approx 68.21^{\circ} \).

06

Calculate maximum height of the secondary rocket above the ground

To find the maximum height, first determine the vertical distance traveled after launch. Use the formula \(v_f^2 = v_i^2 - 2g h\) where \(v_f = 0\) at maximum height, \(v_i = v_{y_{\text{ground}}}=18.10\text{ m/s}\), and \(g = 9.81\text{ m/s}^2\):\(0 = (18.10)^2 - 2(9.81)h\). Solving for \( h \) gives \( h = \frac{(18.10)^2}{2 \times 9.81} \approx 16.69 \text{ m}\). The total height above the ground is therefore the initial height plus the height reached: \( 145 + 16.69 = 161.69 \text{ m} \).

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Key Concepts

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

Horizontal and Vertical Components

When working with projectile motion, it's crucial to analyze the horizontal and vertical components of an object's velocity separately. These components tell us how fast something is moving horizontally (side to side) and vertically (up and down). For a rocket being launched at an angle, we break the motion into these two parts using trigonometry.

If we know the speed of an object and the angle at which it's launched, we can use sine and cosine functions to find these components:

• The horizontal component, denoted as \( v_x \), is calculated by multiplying the initial speed by the cosine of the launch angle. For example, if a rocket is launched at 12 m/s at an angle of 53°, the horizontal component is \( v_x = 12 \cos(53°) \approx 7.21 \text{ m/s} \).
• The vertical component, denoted as \( v_y \), is found by multiplying the initial speed by the sine of the launch angle. Continuing with our example, \( v_y = 12 \sin(53°) \approx 9.60 \text{ m/s} \).

Breaking down velocity into these components helps us predict how far the rocket will travel horizontally and how high it will go vertically, each governed by separate principles of physics.

Relative Velocity

Understanding relative velocity is like understanding perspective. From different viewpoints, the speed and direction of a moving object can appear different. This concept is essential in projectile motion.

Let's say an astronaut in a rocket launches a secondary rocket. To the astronaut, this secondary rocket has a certain speed and direction, broken into its horizontal and vertical components as we previously calculated. However, someone on the ground, like Mission Control, sees both the motion of the primary rocket and the motion of the secondary one.

• The horizontal component of velocity remains the same for both the astronaut and Mission Control, because the primary rocket moves only vertically. Hence, \( v_{x_{ground}} = v_{x_{astronaut}} \approx 7.21 \text{ m/s} \).
• The vertical component, however, differs. To those on the ground, it appears as if the secondary rocket's vertical motion includes the velocity of the primary rocket. Thus, we add the primary rocket's vertical velocity to the vertical component observed by the astronaut: \( v_{y_{ground}} = v_{y_{astronaut}} + 8.50 = 18.10 \text{ m/s} \).

This "adding of velocities" helps us see how and why motion can look different depending on where you stand.

Launch Angle

The launch angle of a projectile significantly affects its trajectory and range. In projectile motion problems, the launch angle tells us how steeply or flatly a projectile is launched, which, in turn, determines how far and how high it will go.

Initially, we know the launch angle as measured by the astronaut. However, since the ground observer sees more motion (the vertical motion added by the primary rocket), they perceive a different launch angle. We use the arctangent function to find this:

• The adjusted launch angle relative to the ground is found using the formula: \( \theta = \tan^{-1}\left(\frac{v_{y_{ground}}}{v_{x_{ground}}}\right) \approx \tan^{-1}\left(\frac{18.10}{7.21}\right) \approx 68.21^\circ \).

This angle suggests a steeper launch in the eyes of someone on the ground compared to what the astronaut observes. Understanding how angles change based on different reference points helps us solve real-world projectile challenges more effectively.