/*! 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 93 When the twin Mars exploration r... [FREE SOLUTION] | 91Ó°ÊÓ

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When the twin Mars exploration rovers, Spirit and Opportunity, set down on the surface of the red planet in January of 2004 their method of landing was both unique and elaborate. After initial braking with retro rockets, the rovers began their long descent through the thin Martian atmosphere on a parachute until they reached an altitude of about \(16.7 \mathrm{m}\). At that point a system of four air bags with six lobes each were inflated, additional retro rocket blasts brought the craft to a virtual standstill, and the rovers detached from their parachutes. After a period of free fall to the surface, with an acceleration of \(3.72 \mathrm{m} / \mathrm{s}^{2}\), the rovers bounced about a dozen times before coming to rest. They then deflated their air bags, righted themselves, and began to explore the surface. Figure \(4-25\) shows a rover with its surrounding cushion of air bags making its first contact with the Martian surface. After a typical first bounce the upward velocity of a rover would be \(9.92 \mathrm{m} / \mathrm{s}\) at an angle of \(75.0^{\circ}\) above the horizontal. Assume this is the case for the problems that follow. How far does a rover travel in the horizontal direction between its first and second bounces? A. \(13.2 \mathrm{m}\) B. \(49.4 \mathrm{m}\) C. \(51.1 \mathrm{m}\) D. \(98.7 \mathrm{m}\)

Short Answer

Expert verified
The rover travels 13.2 meters horizontally, which is option A.

Step by step solution

01

Calculate Horizontal and Vertical Velocity Components

To find out how far the rover travels horizontally, we need to resolve its velocity into horizontal and vertical components. Given that the upward velocity is \(9.92\, \text{m/s}\) at an angle of \(75.0^\circ\):Horizontal velocity, \(v_x = v \cos(\theta) = 9.92 \cos(75.0^\circ)\)Vertical velocity, \(v_y = v \sin(\theta) = 9.92 \sin(75.0^\circ)\)Using a calculator:\[v_x = 9.92 \times \cos(75.0^\circ) = 2.57 \text{ m/s}\]\[v_y = 9.92 \times \sin(75.0^\circ) = 9.57 \text{ m/s}\]
02

Determine the Time of Flight

The time of flight for a projectile launched at a certain vertical velocity can be found using the following equation of motion under uniform acceleration \(g = 3.72 \text{ m/s}^2\) (Martian gravity):\[t = \frac{2v_y}{g}\]Substituting the known values:\[t = \frac{2 \times 9.57}{3.72} = 5.14\, \text{seconds}\]
03

Calculate Horizontal Distance Traveled

The horizontal distance a projectile covers is given by the equation:\[d_x = v_x \times t\]Using the horizontal velocity from Step 1 and the time of flight from Step 2:\[d_x = 2.57 \times 5.14 = 13.2\, \text{meters}\]
04

Choose the Correct Answer

From Step 3, we found that the horizontal distance between the first and second bounces is \(13.2 \text{ meters}\), which corresponds to option A in the given choices.

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Key Concepts

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

Mars Rovers
Mars rovers have been instrumental in exploring Mars, providing a wealth of scientific data to scientists here on Earth. When the twin rovers Spirit and Opportunity landed on Mars in 2004, their mission was to uncover the history of water on the planet. These rovers utilized a unique landing strategy to ensure a safe touchdown on the Martian surface. Instead of using a single method, their landing combined several techniques.

Initially, retro rockets slowed them down, and then a parachute helped them on their descent through the thin atmosphere. To cushion their landing, airbag systems were deployed, allowing them to bounce to a stop rather than landing directly. This method minimized the impact force, protecting the delicate instruments onboard. Once they came to rest, the rovers were able to deflate their airbags, right themselves, and start their exploration mission, roaming several miles on the planet's surface.
Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause the motion. In projectile motion, a special type of motion, kinematics helps us understand how objects move in two-dimensional space. This exercise involving the Mars rovers provides an excellent example of kinematic principles in action.

The way the rovers bounced on the Martian surface following their descent is a classic case of projectile motion. To solve the problem of how far the rovers traveled horizontally, we broke down the velocity into horizontal and vertical components. Using trigonometric functions, we could determine the horizontal and vertical speed at which the rovers moved after their initial impact. This break down is essential because it allows us to calculate the path or trajectory of the rovers accurately.
Martian Gravity
Mars has a gravity of approximately 3.72 meters per second squared, which is about 38% of Earth's gravity. This difference significantly impacts how objects move and fall on Mars compared to Earth. Gravity is the force that makes objects fall to the ground and influences their acceleration.

In this exercise, Martian gravity is crucial as it determines the acceleration of the rovers during their bounce. When solving kinematic equations to find the time of flight or how long it takes for the rovers to bounce, we used Martian gravity rather than Earth's gravity. This adjustment is necessary because the lower gravity would affect the time and distance differently compared to Earth. Understanding the role of Martian gravity helps in predicting the motions of any object on the Martian surface accurately.
Parabolic Trajectory
The path that a projectile follows is known as its trajectory, and it is typically a parabola in shape. In the case of the Mars rovers, when they bounced on the Martian surface, they followed a parabolic trajectory. This means they traveled in an arc-shaped path because of their initial velocity and the force of gravity acting upon them.

When analyzing projectile motion, it's important to recognize that two forces are often at play: the initial force propelling the object, and the constant force of gravity pulling it down. This combination results in a path that is symmetrical and predictable, which is why we could calculate the horizontal distance traveled by the rovers. The understanding of parabolic trajectories is not only important for activities on Mars but also for countless applications on Earth, such as ballistics, sports, and even engineering projects.

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Most popular questions from this chapter

A crow is flying horizontally with a constant speed of \(2.70 \mathrm{m} / \mathrm{s}\) when it releases a clam from its beak (Figure \(4-14) .\) The clam lands on the rocky beach \(2.10 \mathrm{s}\) later. Just before the clam lands, what is (a) its horizontal component of velocity, and (b) its vertical component of velocity? (c) How would your answers to parts (a) and (b) change if the speed of the crow were increased? Explain.

A person flips a coin into the air and it lands on the ground a few feet away. (a) If the person were to perform an identical coin flip on an elevator rising with constant speed, would the coin's time of flight be greater than, less than, or equal to its time of flight when the person was at rest? (b) Choose the best explanation from among the following: I. The floor of the elevator is moving upward, and hence it catches up with the coin in mid flight. II. The coin has the same upward speed as the elevator when it is tossed, and the elevator's speed doesn't change during the coin's flight. III. The coin starts off with a greater upward speed because of the elevator, and hence it reaches a greater height.

A ball rolls off a table and falls \(0.75 \mathrm{m}\) to the floor, landing with a speed of \(4.0 \mathrm{m} / \mathrm{s}\). (a) What is the acceleration of the ball just before it strikes the ground? (b) What was the initial speed of the ball? (c) What initial speed must the ball have if it is to land with a speed of \(5.0 \mathrm{m} / \mathrm{s} ?\)

Pitcher's mounds are raised to compensate for the vertical drop of the ball as it travels a horizontal distance of \(18 \mathrm{m}\) to the catcher. (a) If a pitch is thrown horizontally with an initial speed of \(32 \mathrm{m} / \mathrm{s}\), how far does it drop by the time it reaches the catcher? (b) If the speed of the pitch is increased, does the drop distance increase, decrease, or stay the same? Explain. (c) If this baseball game were to be played on the Moon, would the drop distance increase, decrease, or stay the same? Explain.

An electron in a cathode-ray tube is traveling horizontally at \(2.10 \times 10^{9} \mathrm{cm} / \mathrm{s}\) when deflection plates give it an upward acceleration of \(5.30 \times 10^{17} \mathrm{cm} / \mathrm{s}^{2}\). (a) How long does it take for the electron to cover a horizontal distance of \(6.20 \mathrm{cm} ?\) (b) What is its vertical displacement during this time?

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