The Physics of Everyday Phenomena
8th Edition
ISBN: 9780073513904
Author: W. Thomas Griffith, Juliet Brosing Professor
Publisher: McGraw-Hill Education
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Chapter 3, Problem 32CQ
In the situation pictured in question 31, is the magnitude of the velocity important to the success of the shot? Explain.
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Students have asked these similar questions
The world long jump record is 8.95m. Treated as a projectile, what is the maximum range obtained by a person if he has a take off speed of 9.5m/s? State your assumptions
Two objects are fired into the air. The drawing shows that they reach the same height, but the ranges of their projectile motions are
different. Which one is in the air for the greatest amount of time?
Projectile 1, because it has the smaller range and, hence, the horizontal component of its velocity is smaller than that of
projectile 2.
Both projectiles are in the air for the same amount of time.
Projectile 2, because it has the greater range.
Projectile 1, because it has the smaller initial speed and, therefore, travels more slowly than projectile 2.
The diagram on the right shows the initial velocity of an object. The
magnitude of the initial velocity is vo = 70 m/s. The angle is e = 30
degrees. What are the x and y components of the velocity?
%3D
-60.6 m/s, 35.0 m/s
-35.0 m/s, -60.6 m/s
-60.6 m/s, -35.0 m/s
60.6 m/s, 35.0 m/s
60.6 m/s, -35.0 m/s
Chapter 3 Solutions
The Physics of Everyday Phenomena
Ch. 3 - A small piece of paper is dropped and flutters to...Ch. 3 - The diagram shows the positions at intervals of...Ch. 3 - The diagram shows the positions at intervals of...Ch. 3 - A lead ball and an aluminum ball, each 1 in. in...Ch. 3 - Two identical pieces of paper, one crumpled into a...Ch. 3 - Two identical pieces of paper, one crumpled into a...Ch. 3 - Aristotle stated that heavier objects fall faster...Ch. 3 - A rock is dropped from the top of a diving...Ch. 3 - The graph shows the velocity plotted against time...Ch. 3 - Prob. 10CQ
Ch. 3 - Prob. 11CQCh. 3 - A ball is thrown downward with a large starting...Ch. 3 - A ball thrown straight upward moves initially with...Ch. 3 - A rock is thrown straight upward, reaching a...Ch. 3 - A ball is thrown straight upward and then returns...Ch. 3 - A ball is thrown straight upward and then returns...Ch. 3 - A ball is thrown straight upward. At the very top...Ch. 3 - A ball is thrown straight upward and then returns...Ch. 3 - Prob. 19CQCh. 3 - A ball rolling rapidly along a tabletop rolls off...Ch. 3 - For the two balls in question 20, which, if...Ch. 3 - Is it possible for an object to have a horizontal...Ch. 3 - A ball rolls off a table with a large horizontal...Ch. 3 - A ball rolls off a table with a horizontal...Ch. 3 - An expert marksman aims a high-speed rifle...Ch. 3 - In the diagram, two different trajectories are...Ch. 3 - For either of the trajectories shown in the...Ch. 3 - Assuming that the two trajectories in the diagram...Ch. 3 - A cannonball fired at an angle of 70 to the...Ch. 3 - Will a shot fired from a cannon at a 20 launch...Ch. 3 - The diagram shows a wastebasket placed behind a...Ch. 3 - In the situation pictured in question 31, is the...Ch. 3 - In shooting a free throw in basketball, what is...Ch. 3 - In shooting a basketball from greater than...Ch. 3 - A football quarterback must hit a moving target...Ch. 3 - Prob. 1ECh. 3 - Prob. 2ECh. 3 - Prob. 3ECh. 3 - Prob. 4ECh. 3 - Prob. 5ECh. 3 - Prob. 6ECh. 3 - Prob. 7ECh. 3 - How high above the ground is the ball in exercise...Ch. 3 - At what time does the ball in exercise 7 reach the...Ch. 3 - Prob. 10ECh. 3 - Prob. 11ECh. 3 - Prob. 12ECh. 3 - Prob. 13ECh. 3 - Prob. 14ECh. 3 - Prob. 15ECh. 3 - Prob. 16ECh. 3 - Prob. 1SPCh. 3 - Prob. 2SPCh. 3 - Prob. 3SPCh. 3 - Prob. 4SPCh. 3 - Prob. 5SPCh. 3 - Prob. 6SP
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- 5. A projectile is fired at a 60° angle as shown below. The projectile reaches a maximum height ymax above its initial position. The range of the projectile is 25 m, and the time of flight of the projectile is 3 s. Answer the following questions. 25 a) Find the initial x and y components of the velocity. b) How long does it take the projectile to reach its maximum height? c) What is the final height H of the projectile?arrow_forwardProblem For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) , a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of θ gives the highest maximum height? Solution The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ) a) Let us first find the time it takes for the projectile to reach the maximum height. Using: vfinal-y = vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height is vfinal-y = ? Then, ? = vinitial-y + ayt Substituting the expression of vinitial-y and ay = -g, results to the following: ? = ? - ?t Thus, the time to reach the maximum height is tmax-height = ?/? We will use this time to the equation yfinal - yinitial = vinitial-yt + (1/2)ayt2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = vinitial-yt + (1/2)ayt2 substituting, the vinitial-y…arrow_forward2. Which of the following statements is NOT true regarding projectile motion? a) The range of a projectile is proportional to the square of the initial velocity. c) As mass increases, the deflection of a projectile due to air resistance decreases. b) Air resistance affects the horizontal acceleration of the projectile, but not the vertical. d) The range of a projectile is maximized when launched at an initial angle of 45 degrees.arrow_forward
- Problem For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) , a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of θ gives the highest maximum height? Solution The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ) a) Let us first find the time it takes for the projectile to reach the maximum height.Using: vfinal-y = vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height isvfinal-y =___________ Then, ___________= vinitial-y + aytSubstituting the expression of vinitial-y and ay = -g, results to the following: __________=__________-__________t Thus, the time to reach the maximum height istmax-height = __________ /__________ We will use this time to the equation yfinal - yinitial = vinitial-yt + (1/2)ayt2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax =…arrow_forwardA projectile is launched on the Earth with a certain initial velocity and moves without air resistance. Another projectile is launched with the same initial velocity on the Moon, where the acceleration due to gravity is one-sixth as large. How does the range of the projectile on the Moon compare with that of the projectile on the Earth? (a) It is one-sixth as large, (b) It is the same, (c) It is V5 limes larger, (d) It is 6 times larger, (e) It is 6limes larger.arrow_forwardA ball is fired from a cannon inclined at 360 above the horizontal. The speed with which it leaves the barrel is 200 m/s and the mouth of the cannon is 2 m above ground level. Assume level ground. a) what are the horizontal and vertical components of the initial velocity of the cannon ball ? b) Determine the velocity of the cannon ball at t = 6 seconds. c) How far down range does the ball landarrow_forward
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- Your first task in solving any projectile problem is to determine which case you are dealing with, and that lets you know what equations to use. Case 2 has its own special equations. Case 3 usually involves determining the vertical and horizontal components of the launch velocity. If you know the horizontal velocity and time, R=Vht works for any case of equation.arrow_forwardProblem For a projectile lunched with an initial velocity of vo at an angle of 0 (between 0 and 90°), a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of e gives the highest maximum height? Solution The components of vo are expressed as follows: Vinitial-x = Vocos(0) Vinitial-y = Vosin(e) a) Let us first find the time it takes for the projectile to reach the maximum height. Using: Vfinal-y = Vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height is Vfinal-y %3D Then, Vinitial-y + ayt Substituting the expression of vinitial-y and ay = -g, results to the following: t Thus, the time to reach the maximum height is tmax-height = We will use this time to the equation Yfinal - Yinitial = Vinitial-yt + (1/2)ayt if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = Vinitial-yt + (1/2)ay? substituting, the vinitial-y expression above, results to…arrow_forwardCedar has been trying to get Liam's attention for weeks. He decides to make a paper airplane to hit Liam in the back of his head (and grab his attention). If the paper airplane flies 2.64 m [W 26˚ N] and then is caught by the wind, which causes it to travel 3.21 m [S 12˚ E]. What is the paper airplane's total displacement? (please hand-write it and make it clear)arrow_forward
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