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Chapter 4: Problem 49

An astronaut on Mars kicks a soccer ball at an angle of \(45^{\circ}\) with an initial velocity of \(15 \mathrm{m} / \mathrm{s}\). If the acceleration of gravity on Mars is \(3.7 \mathrm{m} / \mathrm{s}^{2},\) (a) what is the range of the soccer kick on a flat surface? (b) What would be the range of the same kick on the Moon, where gravity is one-sixth that of Earth?

Short Answer

Expert verified
The range of the soccer kick on Mars is approximately 60.81 meters, whereas the range of the same soccer kick on the Moon would be approximately 138.05 meters.

Step by step solution

01

Find the horizontal and vertical components of the initial velocity

Using the angle and the initial velocity given, we can find the horizontal and vertical components of the initial velocity. We will use the trigonometric functions sine and cosine to do this. The horizontal component of the initial velocity (Vx) can be found using the formula: \(Vx = V * cos(\theta)\) The vertical component of the initial velocity (Vy) can be found using the formula: \(Vy = V * sin(\theta)\) - V = 15 m/s (initial velocity) - \(\theta\) = 45掳 (angle)
02

Calculate the components of the initial velocity

Now plug in the values and solve for Vx and Vy. \(Vx = 15 * cos(45^{\circ}) \\ Vx \approx 10.61 \mathrm{m}/\mathrm{s}\) \(Vy = 15 * sin(45^{\circ}) \\ Vy \approx 10.61 \mathrm{m}/\mathrm{s}\)
03

Find the time the soccer ball is in the air

To find the time the soccer ball is in the air, we'll use one of the equations of motion: \(t = \frac{2Vy}{g}\) - t (time) - g = acceleration due to gravity on Mars (3.7 m/ s虏) - Vy = 10.61 m/s (vertical component of the initial velocity)
04

Calculate the time the soccer ball is in the air

Now plug in the values and solve for t. \(t = \frac{2(10.61 \mathrm{m}/\mathrm{s})}{3.7 \mathrm{m}/\mathrm{s}^{2}} \\ t \approx 5.74 \mathrm{s}\)
05

Calculate the range of the soccer kick on Mars

The range of the soccer kick (R) can be calculated using the formula: \(R = Vx * t\) - Vx = 10.61 m/s (horizontal component of the initial velocity) - t = 5.74 s (time the soccer ball is in the air) Now plug in the values and solve for R. \(R = 10.61 \mathrm{m}/\mathrm{s} * 5.74 \mathrm{s} \\ R \approx 60.81 \mathrm{m}\) The range of the soccer kick on Mars is approximately 60.81 meters.
06

Calculate the range of the soccer kick on the Moon

To calculate the range of the same soccer kick on the Moon, we will follow the same steps as before, using the acceleration due to gravity on the Moon, which is one-sixth that of Earth. Earth's gravity is 9.8 m/s虏, so the acceleration due to gravity on the Moon is \(\frac{1}{6} * 9.8 \mathrm{m}/\mathrm{s}^{2} \approx 1.63 \mathrm{m}/\mathrm{s}^{2}\) First, find the time the soccer ball is in the air using the Moon's gravity: \(t_{moon} = \frac{2Vy}{g_{moon}} \\ t_{moon} = \frac{2(10.61 \mathrm{m}/\mathrm{s})}{1.63 \mathrm{m}/\mathrm{s}^{2}} \\ t_{moon} \approx 13.02 \mathrm{s}\) Then, calculate the range of the soccer kick on the Moon: \(R_{moon} = Vx * t_{moon} \\ R_{moon} = 10.61 \mathrm{m}/\mathrm{s} * 13.02 \mathrm{s} \\ R_{moon} \approx 138.05 \mathrm{m}\) The range of the soccer kick on the Moon is approximately 138.05 meters.

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Key Concepts

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

Kinematics
Kinematics is a branch of mechanics that focuses on the motion of objects without considering the forces that cause the motion. When analyzing projectile motion, we break down movement into horizontal and vertical components.
  • Horizontal Motion: This involves constant velocity motion, where the only factor affecting it is the horizontal component of the initial velocity.
  • Vertical Motion: Here, the object's motion is influenced by gravitational acceleration, which causes the object to accelerate downward.
In the exercise above, we use kinematics to calculate how far the soccer ball travels, known as the range. By determining the time it stays airborne and its horizontal velocity, we can find this range using simple kinematic equations.
Gravitational Acceleration
Gravitational acceleration is the acceleration of an object due to the force of gravity. On Earth, this value is approximately \(9.8 \, \text{m/s}^2\), but it varies depending on the celestial body. For instance, gravity on Mars is weaker, at \(3.7 \, \text{m/s}^2\).
  • Impact on Projectile Motion: Gravitational acceleration causes the projectile to follow a parabolic trajectory.
  • Calculating Time in Air: The time an object stays in the air during projectile motion is determined by the equation \(t = \frac{2 \cdot Vy}{g}\), where \(Vy\) is the initial vertical velocity.
For the Moon, with even weaker gravity at \(1.63 \, \text{m/s}^2\), the ball remains in the air longer, leading to a greater range. Thus, gravitational acceleration is crucial for understanding how different environments affect projectile motion.
Trigonometry
Trigonometry is vital in breaking down the components of projectile motion. It helps decipher angles and lengths within the motion's context. The exercise uses trigonometric ratios to find the horizontal and vertical components of the initial velocity.
  • Using Cosine and Sine: For a given launch angle \(\theta\), the horizontal component \(Vx = V \cdot \cos(\theta)\) and the vertical component \(Vy = V \cdot \sin(\theta)\), where \(V\) is the initial speed.
  • 45-Degree Angle: In this specific case, \(\sin(45掳)\) and \(\cos(45掳)\) both equal \(\frac{\sqrt{2}}{2}\), simplifying computations with equal horizontal and vertical velocity components.
By utilizing trigonometry, we accurately decompose the motion's direction, essential for real-world applications like calculating trajectories for different surfaces and gravitational conditions.

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Most popular questions from this chapter

A bullet is shot horizontally from shoulder height \((1.5 \mathrm{m})\) with an initial speed \(200 \mathrm{m} / \mathrm{s}\). (a) How much time elapses before the bullet hits the ground? (b) How far does the bullet travel horizontally?

A runner taking part in the 200 -m dash must run around the end of a track that has a circular arc with a radius of curvature of \(30.0 \mathrm{m}\). The runner starts the race at a constant speed. If she completes the 200 -m dash in 23.2 s and runs at constant speed throughout the race, what is her centripetal acceleration as she runs the curved portion of the track?

The position of a particle changes from 18. The position of a \(\overrightarrow{\mathbf{r}}_{1}=(2.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{cm}\) \(\overrightarrow{\mathbf{r}}_{2}=(-4.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{cm} . \quad\) What is the particle's displacement?

A particle's acceleration is \((4.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2}\). At \(t=0,\) its position and velocity are zero. (a) What are the particle's position and velocity as functions of time? (b) Find the equation of the path of the particle. Draw the \(x-\) and \(y\) -axes and sketch the trajectory of the particle.

A fastball pitcher can throw a baseball at a speed of 40 \(\mathrm{m} / \mathrm{s}(90 \mathrm{mi} / \mathrm{h}) .\) (a) Assuming the pitcher can release the ball 16.7 \(\mathrm{m}\) from home plate so the ball is moving horizontally, how long does it take the ball to reach home plate? (b) How far does the ball drop between the pitcher's hand and home plate?
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