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If a ball is kicked at an angle of 30 degrees such that it has an initial velocity \(v\) , it will travel some distance, \(d_{1}\) before falling back to the ground. Another ball is kicked at an angle of 45 degrees so that it also has an initial velocity of \(v,\) and it travels a distance, \(d_{2}\) , before falling back to the ground. How much farther will the second ball travel before striking the ground? (A) \(\frac{v^{2}}{10}(2-\sqrt{2})\) (B) \(\frac{v^{2}}{10}(2-\sqrt{3})\) (C) \(\frac{v^{2}}{20}(2-\sqrt{2})\) (D) \(\frac{v^{2}}{20}(2-\sqrt{3})\)

Short Answer

Expert verified
The second ball will travel further by a distance of \(\frac{v^2}{20}(2 - \sqrt{3})\), which corresponds to option (D).

Step by step solution

01

Set Up the Projectile Motion Formulas

The range \(d\) of a projectile launched with velocity \(v\) at an angle \(\theta\) can be calculated with the formula: \(d = \frac{v^2}{g} \sin(2\theta)\), where \(g\) is the acceleration due to gravity and is approximately \(10 m/s^2\). Set up the formulas for the two situations: \(d_{1} = \frac{v^2}{10} \sin(60º)\) and \(d_{2} = \frac{v^2}{10} \sin(90º)\).
02

Evaluate Trigonometric Terms for both distances

Evaluate the sine for both angles, remembering that sine of angles in degrees can be solved with the help of a trigonometric values table: \\(1) \$d_{1} = \frac{v^2}{10} \times \sin(60º) = \frac{v^2}{10} \times \frac{\sqrt{3}}{2} = \frac{v^2 \sqrt{3}}{20}\), \\(2) \(d_{2} = \frac{v^2}{10} \times \sin(90º) = \frac{v^2}{10} \times 1 = \frac{v^2}{10}\) .
03

Compute the Difference of Both Distances

Subtract the distance covered by the ball kicked at 30 degrees from the distance covered by the ball kicked at 45 degrees: \(d_{2} - d_{1} = \frac{v^2}{10} - \frac{v^2 \sqrt{3}}{20} = \frac{v^2}{20}(2 - \sqrt{3})\)

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Key Concepts

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

Projectile Motion
Projectile motion is the motion of an object thrown or projected into the air, subject to only the force of gravity. It is a form of two-dimensional motion, meaning it has both horizontal and vertical components. When you kick a ball, it follows a curved path called a trajectory, due to gravity pulling it down.
In the problem, the central concept is calculating how far (the range) a ball travels when kicked at different angles. The range of a projectile can be found using the formula:
  • \(d = \frac{v^2}{g} \sin(2\theta)\), where \(v\) is the initial velocity.
  • \(g\) is the acceleration due to gravity, approximately \(10 m/s^2\).
  • \(\theta\) is the angle of projection.
This formula helps us understand how changes in the launch angle (\(\theta\)) affect the overall distance traveled by the projectile.
In the exercise, two different angles (30° and 45°) yield different ranges, illustrating how angle optimizes distance. It shows that with a constant speed, varying the angle changes the distance the projectile covers, emphasizing the interplay of physics and geometry.
Trigonometry in Physics
Trigonometry is a crucial tool in physics for solving problems involving angles and distances. In projectile motion, it helps dissect the motion into vertical and horizontal components, enabling precise calculation of distances.
In this exercise, we use trigonometric functions like sine to compute the range. The angle of launch influences the trajectory by changing these trigonometric values. For instance:
  • \(\sin(60°) = \frac{\sqrt{3}}{2}\)
  • \(\sin(90°) = 1\)
The sine function helps in determining how much of the initial velocity contributes to the horizontal component of motion, which dictates the range.
This is why for different launch angles, sine values differ, affecting how far the projectile will travel. This exercise shows the applied side of trigonometry, as these values directly impact the distance calculations.
Physics Problems
Physics problems often encompass real-world scenarios like projectile motion. Understanding them builds critical thinking and problem-solving skills, allowing us to see how mathematical concepts apply to tangible situations.
This particular problem is a classic example from the AP Physics 1 Exam that combines the ideas of physics and mathematics. By tackling such problems, students learn to:
  • Identify different forces acting on an object.
  • Apply correct mathematical formulas to describe motion.
  • Develop strategies to dissect the problem into manageable parts.
When approaching physics problems, it’s helpful to:
- Break down the problem into individual components (like angles and velocities).
- Use calculative formulas (like the projectile motion formula) to derive solutions.
Solving exercises like these not only trains students to calculate properly but also enhances their understanding of how physical laws govern the motion of objects in our world.

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Most popular questions from this chapter

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