/*! 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 16 A high school kicker makes a 30.... [FREE SOLUTION] | 91Ó°ÊÓ

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A high school kicker makes a 30.0 -yd field goal attempt (in American football) and hits the crossbar at a height of \(10.0 \mathrm{ft}\). (a) What is the net displacement of the football from the time it leaves the ground until it hits the crossbar? (b) Assuming the football took 2.50 s to hit the crossbar, what was its average velocity? (c) Explain why you cannot determine its average speed from these data.

Short Answer

Expert verified
(a) 90.55 feet, (b) 36.22 feet/s, (c) Trajectory details are missing for average speed.

Step by step solution

01

Convert Units

First, convert the distance from yards to feet, since the height is given in feet. There are 3 feet in a yard, so \(30 \text{ yards} = 90 \text{ feet}\).
02

Determine Net Displacement

The net displacement is a vector that includes both the horizontal and vertical components. The horizontal component is 90 feet (from the goal line to the crossbar), and the vertical component is 10 feet (height of the crossbar). Use the Pythagorean theorem to find the magnitude of the displacement: \[ d = \sqrt{(90)^2 + (10)^2} = \sqrt{8100 + 100} = \sqrt{8200} \approx 90.55 \text{ feet} \].
03

Calculate Average Velocity

Average velocity is the net displacement divided by the time taken. Using the displacement from Step 2 (\(d \approx 90.55 \text{ feet}\)) and the time of 2.50 seconds: \[ v_{avg} = \frac{90.55 \text{ feet}}{2.50 \text{ s}} \approx 36.22 \text{ feet/s} \].
04

Explain Average Speed

Average speed is the total path length divided by time, but we only have the displacement (a straight-line distance). We lack information about the actual curved path of the ball, so we cannot determine the average speed without more details about its trajectory like its maximum height or path arc length.

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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 a form of motion experienced by an object that is launched into the air and moves under the influence of gravity. A key aspect of projectile motion is that the object travels along a curved trajectory called a parabola. For any projectile, such as a football in this exercise, it has two main components:
  • Horizontal component: This is the movement parallel to the Earth's surface. It remains constant, assuming no air resistance.
  • Vertical component: This is the movement perpendicular to the Earth, influenced by gravity, which causes the projectile to accelerate downwards.
When solving projectile motion problems, it's important to analyze these components separately to find overall displacement, velocity, and time of flight. This makes it easier to apply physics concepts and formulas.
Displacement
Displacement in physics refers to the change in position of an object from its starting point to its final position. Unlike distance, which takes into account the entire path traveled, displacement considers only the shortest route between two points—it is a vector quantity. This means it has both magnitude and direction.For the football in this exercise, the net displacement includes both horizontal and vertical movements:
  • Horizontal: The ball travels 90 feet across the field.
  • Vertical: The ball moves up 10 feet to hit the crossbar.
To find the net displacement, use the Pythagorean theorem on these components, as they form a right triangle. The formula \[ d = \sqrt{(horizontal)^2 + (vertical)^2} \]is utilized, resulting in a displacement of approximately 90.55 feet.By understanding displacement, you can better grasp how far out of position something ends up after a journey.
Average Velocity
Average velocity is a measure of the total displacement over the time it takes to make that displacement. It is a vector quantity, meaning it has direction and magnitude, which distinguishes it from average speed (a scalar quantity). In formulaic terms, average velocity is given by:\[ v_{avg} = \frac{\text{displacement}}{\text{time}} \]In this particular case, we had calculated the displacement to be about 90.55 feet, and it took the football 2.50 seconds to reach the crossbar. Thus, the football's average velocity is approximately 36.22 feet per second. It's important to realize that average velocity considers only the starting and finishing points, not the path taken, so it does not fully describe movement where the path isn't straight.
Unit Conversion
Unit conversion is crucial in physics, allowing different units of measurement to be standardized, ensuring consistency and simpler calculations. Converting between units is often necessary, especially when dealing with problems involving vector quantities like displacement or velocity.In this exercise, the field goal attempt was originally given in yards. The challenge required converting this measurement into feet to match the units used for the vertical height of the crossbar. Since there are 3 feet in a yard, the conversion was:\[ 30 \text{ yards} = 30 \times 3 = 90 \text{ feet} \]This step is vital for accurate calculations—mismatched units can lead to incorrect measurements and results. Mastering unit conversions will help you efficiently solve physics problems and handle data correctly.

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Most popular questions from this chapter

Let's investigate a possible vertical landing on Mars that includes two segments: free fall followed by a parachute deployment. Assume the probe is close to the surface, so the Martian acceleration due to gravity is constant at \(3.00 \mathrm{~m} / \mathrm{s}^{2}\). Suppose the lander is initially moving vertically downward at \(200 \mathrm{~m} / \mathrm{s}\) at a height of \(20000 \mathrm{~m}\) above the surface. Neglect air resistance during the free-fall phase. Assume it first free falls for \(8000 \mathrm{~m}\). (The parachute doesn't open until the lander is \(12000 \mathrm{~m}\) from the surface. See \(\mathbf{r}\) Fig. \(2.29 .\) ) (a) Determine the lander's speed at the end of the 8000 -m free-fall drop. (b) At \(12000 \mathrm{~m}\) above the surface, the parachute deploys and the lander immediately begins to slow. If it can survive hitting the surface at speeds of up to \(20.0 \mathrm{~m} / \mathrm{s}\), determine the minimum constant deceleration needed during this phase. (c) What is the total time taken to land from the original height of \(20000 \mathrm{~m} ?\)

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