/*! 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 rookie quarterback throws a fo... [FREE SOLUTION] | 91Ó°ÊÓ

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A rookie quarterback throws a football with an initial upward velocity component of 16.0 \(\mathrm{m} / \mathrm{s}\) and a horizontal velocity component of 20.0 \(\mathrm{m} / \mathrm{s}\) . Ignore air resistance. (a) How much time isrequired for the football to reach the highest point of the trajectory? (b) How high is this point? (c) How much time (after it is thrown) is required for the football to return to its original level? How does this compare with the time calculated in part (a)? (d) How far has the football traveled horizontally during this time? (c) Draw \(x-t\) , \(y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.

Short Answer

Expert verified
(a) 1.63 s (b) 13.06 m (c) 3.26 s, twice of (a) (d) 65.2 m

Step by step solution

01

Determine Time to Reach the Highest Point

To find the time it takes for the football to reach the highest point, consider only the vertical motion. The vertical component of the velocity is \( v_{y} = 16.0 \, \mathrm{m/s} \). At the highest point, the vertical velocity becomes zero. Use the equation for the vertical motion: \( v_{y,f} = v_{y,i} - g \cdot t \). Set \( v_{y,f} = 0 \), and solve for \( t \):\[0 = 16.0 \, \mathrm{m/s} - 9.8 \, \mathrm{m/s}^2 \cdot t \ t = \frac{16.0}{9.8} \, \mathrm{s} \approx 1.63 \, \mathrm{s}. \]
02

Calculate Maximum Height

To find the maximum height, use the formula: \[y = v_{y,i} \cdot t - \frac{1}{2} g \cdot t^2 \]. Substitute the values:\[y = 16.0 \, \mathrm{m/s} \cdot 1.63 \, \mathrm{s} - \frac{1}{2} \cdot 9.8 \, \mathrm{m/s}^2 \cdot (1.63 \, \mathrm{s})^2 \approx 13.06 \, \mathrm{m}. \]
03

Calculate Time to Return to Original Level

The time taken to return to the original level is twice the time taken to reach the highest point because the upward and downward motions are symmetrical. Thus,\[ total \, time = 2 \cdot 1.63 \, \mathrm{s} = 3.26 \, \mathrm{s}. \] The time to descend is equal to the time to ascend.
04

Determine Horizontal Distance Traveled

To calculate how far the football travels horizontally, use the formula: \[x = v_{x} \cdot t_{total} \] where \( v_{x} = 20.0 \, \mathrm{m/s} \) and \( t_{total} = 3.26 \, \mathrm{s} \). Calculate:\[ x = 20.0 \, \mathrm{m/s} \cdot 3.26 \, \mathrm{s} = 65.2 \, \mathrm{m}. \]
05

Graphing (optional step)

To create the graphs for each parameter:1. **x-t graph**: Plot a straight line with a slope of 20.0 m/s, indicating constant horizontal velocity.2. **y-t graph**: Plot a parabola opening downwards, peaking at \(t = 1.63\) s at \(y = 13.06\) m.3. **v_{x}-t graph**: Plot a constant horizontal line at \(20.0\) m/s, showing no change in horizontal velocity.4. **v_{y}-t graph**: Plot a line decreasing from \(16.0\) m/s to 0 m/s at \(t = 1.63\) s, then symmetrically decreasing to \(-16.0\) m/s at \(t = 3.26\) s.

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Key Concepts

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

Kinematics equations
Kinematics equations are fundamental tools in physics used to describe the motion of objects. These equations allow us to predict an object's position, velocity, or acceleration at a given time under uniform acceleration.
For projectile motion, you often deal with two main components: vertical and horizontal. Each component can be analyzed separately using kinematics equations. Commonly used formulas include:
  • Vertical motion: \( v_{y,f} = v_{y,i} - g \cdot t \), where \( v_{y,f} \) is the final vertical velocity, \( v_{y,i} \) is the initial vertical velocity, \( g \) is the acceleration due to gravity, and \( t \) is time.
  • Distance covered: \( y = v_{y,i} \cdot t - \frac{1}{2} g \cdot t^2 \).
  • Horizontal motion: \( x = v_{x} \cdot t \), where \( v_{x} \) is the constant horizontal velocity.
These equations help us determine things like how long it takes for an object to reach the highest point, return to the original level, and the distance traveled. Understanding these formulas is crucial for solving projectile motion problems effectively.
Vertical motion
Vertical motion in projectile problems is all about understanding how an object moves up and down due to gravity. Initially, the object has an upward velocity. At the peak of its trajectory, this vertical velocity becomes zero.
The time taken to reach this point is given by the formula: \( t = \frac{v_{y,i}}{g} \), where \( v_{y,i} \) is the initial vertical velocity. In the problem of the quarterback, it takes about 1.63 seconds to reach the highest point.
To calculate the maximum height the football reaches, you use the equation: \( y = v_{y,i} \cdot t - \frac{1}{2} g \cdot t^2 \).
The maximum height achieved in this specific example is approximately 13.06 meters. This all occurs because gravity is constantly pulling the object downward at a rate of 9.8 m/s².
Horizontal motion
Horizontal motion in projectile problems remains consistent because no external force acts horizontally on the object in such scenarios, assuming air resistance is negligible. This means the horizontal velocity does not change throughout the motion.
For horizontal distance calculation, use the simple formula: \( x = v_{x} \cdot t \), where \( v_{x} \) is the horizontal velocity and \( t \) is the total time the object is in the air.
In the football scenario, with an initial horizontal velocity of 20 m/s and a total air time of 3.26 seconds, the football travels 65.2 meters horizontally.
This consistent motion highlights the independence of vertical and horizontal components, allowing them to be calculated separately.
Graphing motion
Graphing motion provides a visual representation of how different components of motion behave over time.
For projectile motion, we'd typically use four types of graphs, each serving to illustrate specific aspects of the motion:
  • x-t graph: This shows the horizontal position over time. In our scenario, it will be a straight line since the horizontal velocity is constant.
  • y-t graph: This displays the vertical position over time, showing a parabola. It peaks at the maximum height and reflects symmetrical ascent and descent.
  • v_{x}-t graph: This graph remains flat since horizontal velocity does not change.
  • v_{y}-t graph: This plots the vertical velocity, starting positive, reaching zero at the peak, and turning negative as the object descends back.
Understanding these graphs can vastly improve your comprehension of the interplay between different components of motion and how they change over time.

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Most popular questions from this chapter

A 400.0 -m-wide river flows from west to east at 30.0 \(\mathrm{m} / \mathrm{min}\) . Your boat moves at 100.0 \(\mathrm{m} / \mathrm{min}\) relative to the water no matter which direction you point it. To cross this river, you start from a dock at point \(A\) on the south bank. There is a boat landing directly opposite at point \(B\) on the north bank, and also one at point \(C, 75.0 \mathrm{m}\) downstream from \(B(\text { Fig. } 3.53)\) . (a) Where on the north shore will you land if you point your boat perpendicular to the water current, and what distance will you have traveled? (b) If you initially aim your boat directly toward point \(C\) and do not change that bearing relative to the shore, where on the north shore will you land? (c) To reach point \(C :(i)\) at what bearing must you aim your boat, (ii) how long will it take to cross the river, (iii) what distance do you travel, and (iv) and what is the speed of your boat as measured by an observer standing on the river bank?

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