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Chapter 3: Problem 12

A rookie quarterback throws a football with an initial upward velocity component of \(12.0 \mathrm{~m} / \mathrm{s}\) and a horizontal velocity component of \(20.0 \mathrm{~m} / \mathrm{s}\). Ignore air resistance. (a) How much time is required 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? (e) Draw \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.

Short Answer

Expert verified
The football requires 1.22 s to reach the highest point of the trajectory, which is 7.28 m high. The total time for the football to return to its original level is 2.44 s. The football travels a horizontal distance of 48.8 m during this time. The \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion will respectively be a straight line, a parabola opening downwards, a horizontal line, and a straight line with a negative slope according to the aforementioned logic.

Step by step solution

01

Calculate Time to reach the Highest Point

We determine the time required for the football to reach its highest point using the concept that the final vertical velocity at the highest point is 0. We use the equation \( v_f = v_i - g \cdot t \). Here \( v_f = 0 m/s \), \( v_i = 12.0 m/s \), and \( g = 9.8 m/s^2 \). Solving for \( t \), we get \( t = v_i / g \) = 12.0 / 9.8 = 1.22 s.
02

Calculate the Highest Point

We use the equation \( h = v_i t - 0.5gt^2 \) to determine the height at the highest point. Here, \(h\) is the height we are looking for, \(v_i = 12.0 m/s\) is the initial vertical velocity, \(g = 9.8 m/s^2\) is the acceleration due to gravity, and \(t = 1.22 s\) is the time it takes to reach the highest point. Substituting these values, we get \(h = 12.0 \cdot 1.22 - 0.5 \cdot 9.8 \cdot (1.22)^2\) = 7.28 m.
03

Calculate Time for Return

In the absence of air resistance, the time for the football to return to its original level is the same as the time it took to reach the highest point. Therefore, the total time required for the football's complete trip is double the time it took to reach the highest point, which is 2 * 1.22 = 2.44 seconds.
04

Calculate Horizontal Distance

For uniformly linear motion (as the horizontal motion of the football is), we use the formula \(d = v \cdot t\). Here, \(d\) is the horizontal distance, \(v = 20.0 m/s\) is the horizontal velocity, and \(t = 2.44 s\) is the total time for the round trip. Substituting these values, we get \(d = 20.0 \cdot 2.44 = 48.8 m\).
05

Drawing the Graphs

The \(x-t\) graph is a straight line, reflecting uniform motion. The \(y-t\) graph is a parabola opening downwards, indicating a downward acceleration. The \(v_x-t\) graph is a horizontal line, indicating that the horizontal velocity remains constant. The \(v_y-t\) graph is a straight line with a negative slope since the vertical velocity decreases linearly due to the constant acceleration - gravity. The heights of these graphs will be determined by the initial velocities and the time it takes for the football to complete its round trip.

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Key Concepts

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

Kinematics
Kinematics is the study of motion without considering the forces that cause it. It's all about describing how an object moves. In a projectile motion scenario like the football throw, we break down motion into two components: horizontal and vertical.
  • Horizontal motion happens with constant velocity
  • Vertical motion is influenced by gravity
In the case of the quarterback's football, kinematics helps us predict the path taken by splitting its velocity into horizontal and vertical components. With a horizontal component of 20 m/s, the motion remains steady and uninfluenced by gravity. Vertically, an upward initial velocity of 12 m/s will be acted upon by the gravitational pull, which is constant at 9.8 m/s虏.
By applying kinematic equations, we can solve for unknowns like time, distance, and acceleration in each direction. Understanding these principles ensures that we can predict how the football behaves throughout its throw鈥攆rom launch to landing.
Free Fall
Free fall is when an object moves solely under the influence of gravity. In projectile motion, the vertical component behaves like a free-falling object. The effect of gravity means that the vertical speed of the football decreases as it ascends until it pauses momentarily at the peak.

Once at the peak, the football begins to descend, speeding up under gravitational acceleration until it returns to the same vertical level. This happens with increasing ease because gravity acts consistently on any object in free fall.
  • Upward motion: slows down
  • Peak: momentarily stops
  • Downward motion: speeds up
Free fall shows how gravity's pull impacts the football, enabling calculation of how high it goes (7.28 m) and the time taken (1.22 s) to reach this peak.
Graphs of Motion
Graphs offer a visual representation of how the football's motion changes over time. Analyzing these graphs can make kinematic relationships clearer.
  • Position-time graph (x-t): Horizontal motion appears as a straight line representing constant velocity.
  • Position-time graph (y-t): This shows a parabolic shape due to the quadratic nature of free fall.
  • Velocity-time graph (v_x-t): Here, the line is horizontal because the horizontal velocity is constant at 20 m/s.
  • Velocity-time graph (v_y-t): This will slope downwards, indicating the vertical velocity's decrease from 12 m/s to 0 m/s at the peak, then back to -12 m/s when it returns to the starting level.
Graphs like these are crucial in understanding the characteristics of motion, making it easier to interpret the effects of uniform acceleration and constant speeds during projectile motion.

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Most popular questions from this chapter

The earth has a radius of \(6380 \mathrm{~km}\) and turns around once on its axis in \(24 \mathrm{~h}\). (a) What is the radial acceleration of an object at the earth's equator? Give your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and as a fraction of \(g .\) (b) If \(a_{\mathrm{rad}}\) at the equator is greater than \(g\), objects will fly off the earth's surface and into space. (We'll see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determine the width of the river and the current speed (the speed of the water relative to the earth). You have a small boat with an outboard motor. By measuring the time it takes to cross a pond where the water isn't flowing, you have calibrated the throttle settings to the speed of the boat in still water. You set the throttle so that the speed of the boat relative to the river is a constant \(6.00 \mathrm{~m} / \mathrm{s}\). Traveling due north across the river, you reach the opposite bank in \(20.1 \mathrm{~s}\). For the return trip, you change the throttle setting so that the speed of the boat relative to the water is \(9.00 \mathrm{~m} / \mathrm{s}\). You travel due south from one bank to the other and cross the river in \(11.2 \mathrm{~s}\). (a) How wide is the river, and what is the current speed? (b) With the throttle set so that the speed of the boat relative to the water is \(6.00 \mathrm{~m} / \mathrm{s},\) what is the shortest time in which you could cross the river, and where on the far bank would you land?

An object moves in a horizontal circle at constant speed \(v\) (in units of \(\mathrm{m} / \mathrm{s}) .\) It takes the object \(T\) seconds to complete one revolution. Derive an expression that gives the radial acceleration of the ball in terms of \(v\) and \(T,\) but not \(r .\) (a) If the speed doubles, by what factor must the period \(T\) change if \(a_{\text {rad }}\) is to remain unchanged? (b) If the radius doubles, by what factor must the period change to keep \(a_{\text {rad }}\) the same?

A jet plane is flying at a constant altitude. At time \(t_{1}=0,\) it has components of velocity \(v_{x}=90 \mathrm{~m} / \mathrm{s}, v_{y}=110 \mathrm{~m} / \mathrm{s} .\) At time \(t_{2}=30.0 \mathrm{~s}\) the components are \(v_{x}=-170 \mathrm{~m} / \mathrm{s}, v_{y}=40 \mathrm{~m} / \mathrm{s}\). (a) Sketch the velocity vectors at \(t_{1}\) and \(t_{2} .\) How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration.

A small ball is attached to the lower end of a 0.800 -m-long string, and the other end of the string is tied to a horizontal rod. The string makes a constant angle of \(37.0^{\circ}\) with the vertical as the ball moves at a constant speed in a horizontal circle. If it takes the ball \(0.600 \mathrm{~s}\) to complete one revolution, what is the magnitude of the radial acceleration of the ball?
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