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When it is 145 \(\mathrm{m}\) above the ground, a rocket traveling vertìcally 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. Air resistance is too small to worry about. (a) Just as the secondary rocket is lunched, 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?

Short Answer

Expert verified
(a) Relative to astronaut: horizontal 7.21 m/s, vertical 9.60 m/s. Relative to ground: horizontal 7.21 m/s, vertical 18.10 m/s. (b) Total speed: 19.6 m/s, angle 68.6°. (c) Max height: 271 m above ground.

Step by step solution

01

Understand the Scenario

A rocket is moving vertically at 8.50 m/s. A secondary rocket is launched from this moving rocket, traveling at 12.0 m/s at an angle of 53.0° relative to the horizontal coordinate system of the initial rocket.
02

Relative to the Astronaut - Horizontal and Vertical Velocity

For the secondary rocket, break down its velocity of 12.0 m/s into horizontal and vertical components. Relative to the astronaut: - Horizontal: \( v_{x, astronaut} = 12.0 \cos(53.0^{\circ}) \)- Vertical: \( v_{y, astronaut} = 12.0 \sin(53.0^{\circ}) \)
03

Calculate Horizontal and Vertical Components Relative to Ground

To find the speed of the secondary rocket relative to the ground, we add the initial vertical speed of the main rocket to the secondary rocket’s vertical component:- Horizontal (same from astronaut's frame): \( v_{x, ground} = 12.0 \cos(53.0^{\circ}) \)- Vertical: \( v_{y, ground} = 8.50 + 12.0 \sin(53.0^{\circ}) \)
04

Combine Components for Total Speed and Launch Angle Relative to Ground

Using Pythagorean theorem and trigonometry, calculate:- Total speed \( v_{total, ground} = \sqrt{v_{x, ground}^2 + v_{y, ground}^2} \)- Launch angle \( \theta = \tan^{-1}\left(\frac{v_{y, ground}}{v_{x, ground}}\right) \)
05

Find Maximum Height Above Ground

Use kinematic equations for vertical motion to determine the maximum height reached. The formula used is:\[ v_{y, ground}^2 = 2gh \]or rearrange to find 'h'. With initial height, total height \( = 145 + h \) above the ground.

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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 a powerful tool to describe the motion of objects under constant acceleration. They help us predict different aspects of motion like speed, time, and distance. For example, to find out how high an object will rise or how long it stays in the air, we often rely on these equations.

In projectile motion, we use kinematic equations to understand the motion in both vertical and horizontal directions separately. Some common equations include:
  • Velocity: \( v = u + at \)
  • Displacement: \( s = ut + \frac{1}{2}at^2 \)
  • Velocity squared: \( v^2 = u^2 + 2as \)
Here, \( v \) is final velocity, \( u \) is initial velocity, \( a \) is acceleration, and \( t \) is time. Displacement (\( s \)) is the change in position, considering the direction.

In our exercise, these formulas help calculate how high the secondary rocket reaches above the ground, given its initial vertical velocity and the acceleration due solely to gravity.
Relative Velocity
Relative velocity describes how the speed of an object is perceived differently depending on the observer's point of view. In the context of our exercise, this concept is key to understanding how the speeds of the rockets are measured differently by the astronaut in the main rocket and by Mission Control on the ground.

When calculating relative velocities, it's essential to account for both the motion of the observer and the speed of the moving object. For example:
  • Relative to the astronaut, the secondary rocket's initial velocity is purely based on its launch speed and angle.
  • Relative to the ground, the main rocket's speed adds to the vertical motion of the secondary rocket, adjusting its perceived speed.
Grasping this concept allows us to understand the different perspectives and the importance of adding velocities accurately based on the reference frames involved.
Launch Angle
The launch angle is the angle at which an object is projected into the air, measured relative to the horizontal. It's a crucial factor in determining the range and height the object reaches.

In our scenario, the astronaut launches the secondary rocket at an angle of \(53.0^{\circ}\). This specific angle affects how the velocity of the rocket is split into horizontal and vertical components. Knowing how to handle the launch angle is pivotal to analyze the complete projectile motion, as this angle directly influences:
  • Horizontal distance traveled.
  • Maximum height achieved.
A larger angle tends to increase the height an object will reach, while a smaller angle might favor the distance covered. In this way, understanding angles helps in predicting the rocket's behavior after launch.
Vertical and Horizontal Components
Projectile motion often involves splitting a velocity into vertical and horizontal components. This separation is important because it allows us to treat the motion in two dimensions independently.

For instance, the secondary rocket's velocity can be divided into:
  • Horizontal component: This is unaffected by gravity and remains constant since no horizontal forces are acting on the projectile besides air resistance, which is negligible here.
  • Vertical component: This is influenced by gravity, which will decelerate the rocket until it reaches its peak and accelerate it back downwards.
Using trigonometric functions, we find these components from the launch speed and angle as follows:
  • Horizontal velocity: \( v_{x} = v \cos(\theta) \)
  • Vertical velocity: \( v_{y} = v \sin(\theta) \)
This method helps calculate the secondary rocket's different perceived speeds, both by an astronaut in the moving rocket and an observer on the ground, for precise analysis of its trajectory.

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Most popular questions from this chapter

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Leaping the River I. A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle (Fig. 3.51\() .\) The takeoff ramp was inclined at \(53.0^{\circ}\) , the river was 40.0 \(\mathrm{m}\) wide, and the far bank was 15.0 \(\mathrm{m}\) lower than the top of the ramp. The river itself was 100 \(\mathrm{m}\) below the ramp. You can ignore air resistance. (a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? (b) If his speed was only half the value found in (a), where did he land?

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