Two-Dimensional and Projectile Motion [WEEK 2]
James H Dann, Ph.D. James Dann, Ph.D.
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AUTHORS James H Dann, Ph.D. James Dann, Ph.D. CONTRIBUTORS Antonio De Jesus López Chris Addiego
CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2013 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: September 8, 2013
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Chapter 1. Two-Dimensional and Projectile Motion [WEEK 2]
C HAPTER
1
Two-Dimensional and Projectile Motion [WEEK 2]
C HAPTER O UTLINE 1.1
Vectors
1.2
Velocity
1.3
Projectile Motion
1.4
Projectile Motion Problem Solving
1.5
References
Introduction - Two-Dimensional and Projectile Motion [WEEK 2] The Big Idea
In this chapter, we explore the motion of projectiles under the influence of gravity — fired cannonballs, thrown basketballs, and other objects that have no way of propelling themselves and do not experience significant air resistance. We know that vectors can be separated into components (see first lesson); if they are separated into perpendicular components the motion along each component can be treated independently. This is the insight that allows us to solve two dimensional projectile motion problems: we break any initial velocity vector into a component parallel to the ground and a component perpendicular to it. The force of gravity — which will be explained in more detail later — accelerates any object near the surface of the earth toward its center at a rate of g = 9.8m/s2 . This acceleration is in the direction perpendicular to the surface of the earth, conventionally labeled y. Since in projectile motion under the sole influence of gravity any acceleration the object experiences is in the y direction, its horizontal, or x, velocity remains constant throughout its flight (at least in the absence of air resistance, which we ignore for the time being). To solve two dimensional motion problems, we apply the kinematics equations of one-dimensional motion to each of the two directions. In the y direction, we can use the uniform acceleration equations to solve for time in flight. Using this time, we can find how far the object traveled in the x direction also.
1
1.1. Vectors
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1.1 Vectors In order to solve two dimensional problems it is necessary to break all vectors into their x and y components. Different dimensions do not ’talk’ to each other. Thus one must use the equations of motion once for the x-direction and once for the y-direction. For example, when working with the x-direction, one only includes the x-component values of the vectors in the calculations. Note that if an object is ’launched horizontally’, then the full value is in the x-direction and there is no component in the y-direction. In order to solve two dimensional problems it is necessary to break all vectors into their x and y components. Different dimensions do not ’talk’ to each other. Thus one must use the equations of motion once for the x-direction and once for the y-direction. For example, when working with the x-direction, one only includes the x-component values of the vectors in the calculations. Note that if an object is ’launched horizontally’, then the full value is in the x-direction and there is no component in the y-direction.
Key Equations Vectors
The first new concept introduced here is that of a vector: a scalar magnitude with a direction. In a sense, we are almost as good at natural vector manipulation as we are at adding numbers. Consider, for instance, throwing a ball to a friend standing some distance away. To perform an accurate throw, one has to figure out both where to throw and how hard. We can represent this concept graphically with an arrow: it has an obvious direction, and its length can represent the distance the ball will travel in a given time. Such a vector (an arrow between the original and final location of an object) is called a displacement:
Vector Components
From the above examples, it should be clear that two vectors add to make another vector. Sometimes, the opposite operation is useful: we often want to represent a vector as the sum of two other vectors. This is called breaking a vector into its components. When vectors point along the same line, they essentially add as scalars. If we break vectors into components along the same lines, we can add them by adding their components. The lines we pick to break our vectors into components along are often called a basis. Any basis will work in the way described above, but we usually break vectors into perpendicular components, since it will frequently allow us to use the Pythagorean theorem in time-saving ways. Specifically, we usually use the x and y axes as our basis, and therefore break vectors into what we call their x and y components: 2
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Chapter 1. Two-Dimensional and Projectile Motion [WEEK 2]
A final reason for breaking vectors into perpendicular components is that they are in a sense independent: adding vectors along a component perpendicular to an original component one will never change the original component, just like changing the y-coordinate of a point can never change its x-coordinate.
Break the Initial Velocity into its Components
Example 1
A tennis ball is launched 32◦ above the horizontal at a speed of 7.0 m/s. What are the horizontal and vertical velocity components? Question: vx and vy = ? [m/s] Given: v = 7.0 m/s θ = 32◦ 3
1.1. Vectors
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Equation: vx = v cos θ
vy = v sin θ
Plug n’ Chug: vx = v cos θ = (7.0 m/s) cos(32◦ ) = 5.9 m/s vy = v sin θ = (7.0 m/s) sin(32◦ ) = 3.7 m/s Answer: 5.9 m/s, 3.7 m/s.
Watch this Explanation
MEDIA Click image to the left for more content.
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Time for Practice
1. Find the missing legs or angles of the triangles shown.
2. Draw in the x− and y−velocity components for each dot along the path of the cannonball. The first one is done for you. 4
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Chapter 1. Two-Dimensional and Projectile Motion [WEEK 2]
5
1.2. Velocity
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1.2 Velocity Students will learn how analyze and solve problems in two dimensions containing velocity vectors but no acceleration. Students will learn how analyze and solve problems in two dimensions containing velocity vectors but no acceleration.
Key Equations
vx = v cos θ, where θ is the angle between the velocity vector and the horizontal. vy = v sin θ, where θ is the angle between the velocity vector and the horizontal. ∆x = vxt ∆y = vyt
Guidance
The only equation you need is that displacement in a certain direction equals the component of velocity in that direction multiplied by the time it takes. You’ll use this once for the x-direction and once for the y-direction and solve for what is asked.
Example 1
Question: If a river is flowing north at 2 m/s and you swim straight across (i.e. east) at 1.5 m/s, how far up shore will you be from your starting point once you reach the other side? The river is 9 m wide. Answer: First solve for the time it takes you to reach the other side. Let’s let north be the y-direction and the direction across the river be the x-direction. ∆x = vxt 9m = 1.5m/s × t thus, t = 6s Now, use the time you are in the water to find how far the river has carried you north. ∆y = vyt ∆y = 2m/s × 6s ∆y = 12m
Watch this Explanation
6
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Chapter 1. Two-Dimensional and Projectile Motion [WEEK 2]
MEDIA Click image to the left for more content.
Simulation
• http://simulations.ck12.org/RelativeVelocity/ Time for Practice
1. If a river is flowing south at 4 m/s and you swim straight across (i.e. east) at 2 m/s; admittedly, you’re going to drift a bit south. That said, calculate that distance that you drifted south from your starting point. The river is 16 m wide. 2. If a river is flowing south at 3 m/s and you swim at an angle of 30 degrees north of directly east at 1 m/s, how far did you drift up or down stream from your starting point once you reach the other side? The river is 10 m wide. 3. If a river is flowing north at 2 m/s and you can swim at 4 m/s, what angle should you swim at such that you arrive directly across the river (i.e. no drift north or south from starting point on other side)? The river is 10 m wide. 4. If a river is flowing south at 5 m/s and you can swim at 4 m/s maximum, is it possible to arrive directly across? Why or why not? Answers to Selected Problems
1. 2. 3. 4.
32 m 28.9 m south of starting point 30 degrees No, even if you swim directly north the river will still take you south at 1 m/s
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1.3. Projectile Motion
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1.3 Projectile Motion • Solve problems involving objects experiencing uniform horizontal motion and accelerated vertical motion • Understand relative motion in a vertical plane Vocabulary
Free fall The condition of acceleration which is due only to gravity. An object in free fall is not being held up, pushed, or pulled by anything except its own weight. Though objects moving in air experience some force from air resistance, this is sometimes small enough that it can be ignored and the object is considered to be in free fall. Projectile motion Projectile motion is a form of motion where an object (called a projectile) is thrown near the earth’s surface with some horizontal component to its velocity. The projectile moves along a curved path under the action of gravity. The path followed by a projectile is called its trajectory. Projectile motion is motion in two directions. In the vertical direction, the motion is accelerated motion and in the horizontal direction, the motion is constant velocity motion. Instantaneous velocity of a projectile Instantaneous velocity is the velocity of an object at one instant during its motion. In the case of a projectile, the instantaneous velocity vector would be the resultant of a constant velocity horizontal motion and an accelerated velocity vertical motion. Range A projectile launched with specific initial conditions will travel a predictable horizontal displacement before striking the ground. This distance is referred to as the projectile’s range. Equations
y=
vi +v f 2
t
x f = (v cos θ)t + xi y f = 12 gt 2 + (v sin θ)t + yi vx = v cos θ vy = v sin θ Introduction
“Independence of Motion along Each Dimension” looks at the special case of throwing an object in a purely horizontal direction. Projectile motion is the general case of throwing an object in any direction, from sliding an object off a desk to kicking a soccer ball as shown in the Figure below. http://demonstrations.wolfram.com/JumpingOverRowOfParkedCars/ 8
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Chapter 1. Two-Dimensional and Projectile Motion [WEEK 2]
FIGURE 1.1 Author:
Soccer player:
Image copy-
right Mandy Godbehear, 2012; gram:
Dia-
CK-12 Foundation - Christo-
pher Auyeung License:
Used under
license from Shutterstock.com Source: http://www.shutterstock.com
Finding instantaneous velocity
Let’s look back at the case of a dart fired out from a dart gun, using vector mechanics. Consider the trajectory of the dart sometime between pulling the trigger and striking the ground. The diagram below is a representation of the instantaneous velocity components of the dart.
FIGURE 1.2 Author: CK-12 Foundation - Raymond Chou License: CC-BY-NC-SA 3.0
Using only one-dimensional mechanics, we found that in the y-direction, it takes 0.55 seconds for the dart to hit the floor from a height of 1.5 meters. If the dart traveled a horizontal distance of 6.0 m in 0.55 seconds, then its horizontal component of velocity is therefore vx =
6.0m = 10.9 m/s 0.55s
. What is the vector velocity at time, t = 0.25s ? The x and y velocity components represent the legs of the right triangle (see Figure above) and the hypotenuse represents the instantaneous resultant velocity of the projectile. Using the Pythagorean Theorem, vx 2 + vy 2 = v2 , gives the magnitude of the instantaneous velocity of the projectile once the square root is taken. The x velocity is constant, so we know it is 10.9 m/s at all times. For the y velocity, we know from one-dimensional motion that v f = at + vi . The acceleration from gravity is constant, giving: (−10m/s2 )(0.25s) + 0m/s = −2.5 m/s. Now that both legs of the right triangle are known, we can apply the Pythagorean Theorem to solve for the instantaneous speed (the hypotenuse of the right triangle) at time, t = 0.25 s. 9
1.3. Projectile Motion
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q q 2 2 v = vx + vy = (10.9m/s)2 + (2.5m/s)2 = 11.2 m/s If all that was asked for in this problem was the instantaneous speed of the dart at 0.25 s we would be done. However, you may be asked for the dart’s instantaneous velocity at t = 0.25 s. In that case, you need to determine the instantaneous direction of the dart as well (recall that velocity is a vector quantity, and as such, has a magnitude and a direction). Finding the instantaneous direction of the velocity of the dart at t = 0.25 s: Using the tangent function is the most efficient method. The angle that we need to find is somewhat arbitrary; after all, other than the right angle, the triangle has two perfectly good angles to choose from. The angle usually preferred, however, is measured from x−direction. Using the inverse tangent relationship which can be found on your calculator, we have: 2.5m/s −1 ◦ tan θ = 10.9 m/s ; tan θ = 12.7 If you are asked to represent the vector using the trigonometric definition of angle measurement, than either −12.7◦ or 347.3◦ would do. The final vector result can be stated as (11.2m/s, −12.7◦ ). Components of projectile motion
In the problem that follows we imagine the projectile launched from the ground with an angular elevation between 0 and 90 degrees, with an initial x−component of velocity of +30 m/s and an initial y−component of velocity of +40 m/s.
FIGURE 1.3 Author: CK-12 Foundation - Christopher Auyeung License: CC-BY-NC-SA 3.0
Some typical questions that can be asked in such a situation are: 1. 2. 3. 4.
What is the time the projectile takes to reach the highest position above the ground? What is the projectile’s highest position above the ground? What is the velocity of the projectile at its highest position above the ground? What is the range of the projectile?
1. In order to answer the first question, let’s consider what determines the amount of time the projectile remains airborne. A velocity of 30 m/s in the x−direction, does not affect the time in the air; no more so than the gun or dart’s airborne time was affected by their horizontal motion. The vertical component of motion, however, must determine how much time the projectile spends airborne since only the vertical motion is subject to gravity. If we imagine just the vertical component of +40 m/s we can quickly estimate (if we make the approximation that -9.8 is close enough to -10) the amount of time it takes the projectile to reach its highest position (peak); it’s about 4 40 seconds, or if we need more accuracy; 9.8 = 4.08 s = 4.1 s. Remember that the projectile loses 9.8 m/s of velocity every second it ascends. Thus, 10
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Chapter 1. Two-Dimensional and Projectile Motion [WEEK 2]
1. Answer: t = 4.1 s v +v 2. Answer: Once again we can use x = vt that is, y = i 2 f t, as long as we understand that v represents the average velocity for the ascent part of the projectile’s motion. Since the projectile began with an initial velocity in the y−direction of 40 m/s and has a final velocity of 0 m/s, at the top of its motion in the y−direction, its average velocity is (40+0) = 20 m/s. Its highest position above the ground is therefore, (20 m/s)(4.08) = 81.6 = 82 m. 2 3. Answer: It is often incorrectly thought that the velocity of the projectile at its peak position is zero. It must be remembered that until the projectile hits the ground, it is always moving horizontally. Though the vertical velocity is zero at the top of its motion, the horizontal velocity is still +30 m/s. Therefore, the velocity of the projectile at its highest position above the ground is +30 m/s (the + indicating the projectile is moving with a velocity of 30 m/s to the right). 4. Answer: The range of the projectile is determined by finding the total time the projectile is airborne and multiplying that time by its speed in the horizontal direction (x = vt). The projectile’s motion is symmetric in the absence of air friction so the total time for the trip is twice 4.08 s or 8.16 s. The range is, therefore, (30)(8.16) = 244.8 m = 240 m. Of course, instead of using a bit of physical reasoning and simple arithmetic, we could have used more sophisticated equations to answer all these questions, but why make things more complicated than they have to be? Before leaving this problem, let’s alter the given information, in order to see another way the problem could have been stated. A projectile is launched from the ground at an angle of 53.13 degrees with a speed of 50 m/s. Find the answers to questions 1-4 above. Our claim is that both problems are, in fact, identical and have the same answers. If we’re not given the horizontal and vertical components of the velocity, it is good policy to find them before trying to solve a problem of this sort. The x−component (horizontal) is: vx = v cos θ → 50 cos 53.13◦ = 30 m/s The y−component (vertical) is: vyi = v sin θ → 50 sin 53.13◦ = 39.999 → 40 m/s, Since we see that the components are identical to the original problem, we can solve the problem the same way. http://demonstrations.wolfram.com/ThrowingABaseballFromTheOutfieldToHomePlate/ An interesting aside: If at 4.1 s, (the time when the projectile has reached its maximum height) we pretend to erase the first half of the projectile’s motion and label the projectile at its peak position with a vector (rather an a component of a vector) pointing to the right with a speed of 30 m/s, the remaining path of the projectile would be similar to the bullet and dart of the previous section (3.3 “A Special case of Projectile Motion”) and the problem would be solved in the same manner as both bullet and dart problems. See Figure below.
FIGURE 1.4 Author: CK-12 Foundation - Christopher Auyeung License: CC-BY-NC-SA 3.0
Let’s take a trip! 11
1.3. Projectile Motion
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FIGURE 1.5 The package remains under the plane, but people on ground see the package’s descent as a parabola.
Author:
Im-
age copyright Konstantin Yolshin, 2012; modified by CK-12 Foundation - Christopher Auyeung License:
Used under
license form Shutterstock.com Source: http://www.shutterstock.com
Check your understanding
Imagine a small airplane flying at a constant elevation of 80 m over your school soccer field, with a velocity of 50 m/s, (about to 110 mph), eastward; see Figure above. Ignore air resistance. 1. What would the motion of the package look like to an observer on the ground? Answer: An observer on the ground would see the package fall in a parabolic arc, as if it had been projected horizontally with a speed of 50 m/s; the same motion performed by the bullet and dart; (though the package would remain directly below the airplane). 2. What would the motion of the package look like to an observer on the plane? Answer: The package would appear to fall straight down, since the package and the airplane both have the horizontal velocity of 50.0 m/s. 3. The plane releases the package when it flies over a target marked A on the ground, will the package land on A? Answer: No, it will not, since, as the package falls, it is moving at the horizontal velocity 50 m/s. 4. How far from point A does the package land? Answer: If we know how much time the package spends traveling at 50.0 m/s after it is released, we can use x = vt to solve the problem. The time the package remains in the air is found using our one-dimensional equation: y f = 12 at 2 + viyt + yi where a = g = −10 m/s2 , viy = 0 m/s (at the instant the package is released it has no velocity in the y−direction), y f = 0 and yi = 100 m; therefore, 21 (−10m/s2 )t 2 + 0 + 80m = 0. Solving for t, we find t = 4.0 s. Thus x = (50.0m/s)(4.0s) = 200m. The package fell 200 m past target A. In order for the pilot to hit target A, the package must be released 200m before reaching the target, or 4.0 seconds before reaching target A. See Figure below Deriving some general results from the projectile equations
From the problems that we discussed, it should be reasonably clear that the equations we derived for one-dimensional kinematics can be used, with caution, for 2-dimentional kinematics. The only difference is that care must be taken in correctly identifying velocity components. Recall that if velocity is stated as V at an angle of θ[U+F02C] then 12
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Chapter 1. Two-Dimensional and Projectile Motion [WEEK 2]
FIGURE 1.6 Author: Image copyright Konstantin Yolshin, 2012; modified by CK-12 Foundation - Christopher Auyeung License: Used under license form Shutterstock.com Source: http://www.shutterstock.com
the initial x and y components of the velocity are, v cos θ and v sin θ, respectively. With this in mind, we rewrite our kinematic equations as follows: 1. x − direction : x f = R = (v cos θ)t + xi 2. y − direction : y f = 12 gt 2 + (v sin θ)t + yi The other equations are treated the same way, using the initial velocity in the y−direction: 3. vy f = gt + v sin θ 4. vy f 2 = (v sin θ)2 + 2g∆y 5. vave =
v sin θ+v f y 2
Special cases
What is the maximum range, R, of a projectile if yi = y f = 0? setting yi and y f = 0 in equation 2, and factoring out t, we have: t 12 gt + v sin θ = 0 There are two solutions: t = 0 and t =
−2v sin θ g
The trivial condition is satisfied at launch (t = 0, yi = 0) and the nontrivial condition is satisfied at landing t = If t =
−2 sin θ g ,yf
−2v sin θ g
is substituted into equation 1 : R = (v cos θ)t 2 We have: R = (v cos θ) −2vgsin θ = −2v sing θ cos θ Using the trigonometric identity 2 sin θ cos θ = sin 2θ, we have: R=
−v2 sin 2θ g
We can extract a useful result from the range equation. Assuming that g is constant, the range is a function of the V and θ. Let’s consider the angle. Since the maximum value of the sine is 1.0 then whatever angle makes sin 2θ = 1, will also maximize the range. Since sin 90◦ = 1, 2θ = 90◦ and the angle which produces the greatest range is 45◦ . The range continually increases with increasing velocity so there is no “interesting” information we can get out of 13
=0 .
1.3. Projectile Motion
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V , except to say that V must be is at least some minimum value if the intended target is to be reached. Since 45◦ gives the greatest range, what kind of range will smaller and larger angles than 45◦ give? Forty-five degrees is an optimum condition and we notice that for θ = 30◦ and θ = 60◦ (equally distributed about 45◦ we have 2θ = 2(30◦ ) = 60◦ , and 2θ = 2(60◦ ) = 120◦ which gives sin 60◦ = sin 120◦ . The implication is that the range is the same for angles (45◦ + θ) and (45◦ − θ). This condition can be readily proven with a bit of trigonometry: Is sin(2(45◦ + θ)) = sin(2(45◦ − θ)) an identity? The distribution gives sin(90◦ + 2θ) = sin(90◦ − 2θ), which after expansion gives: cos(2θ) = cos(2θ) and confirms the statement is an identity. Lastly, we consider expressing y as a function of x, rather than t. By solving equation 1 for t and substituting the result into equation 2, we have, after recalling tan θ =
1 (∆x)2 + ∆x tan θ + yi yf = g 2 (v cos θ)2 If Xi = 0 then ∆x above can be replaced with X, or R.
14
sin θ cos θ ,
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Chapter 1. Two-Dimensional and Projectile Motion [WEEK 2]
1.4 Projectile Motion Problem Solving Students will learn how to use the equations of motion in two dimensions in order to solve problems for projectiles. It is necessary to understand how to break a vector into its x and y components. Students will learn how to use the equations of motion in two dimensions in order to solve problems for projectiles. It is necessary to understand how to break a vector into its x and y components.
Key Equations Break the Initial Velocity Vector into its Components
Apply the Kinematics Equations
Vertical Direction
Horizontal Direction
y(t) = yi + viyt − 21 gt 2
x(t) = xi + vixt
vy (t) = viy − gt
vx (t) = vix
2
2
vy = v0y − 2g(∆y) ay = −g = −9.8m/s2 ≈ −10m/s2
ax = 0
15
1.4. Projectile Motion Problem Solving
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Guidance
• To work these problems, separate the “Big Three” equations into two sets: one for the vertical direction, and one for the horizontal. Keep them separate. • The only variable that can go into both sets of equations is time; use time to communicate between the x and y components of the object’s motion.
Example 1
CSI discovers a car at the bottom of a 72 m cliff. How fast was the car going if it landed 22m horizontally from the cliff’s edge? (Note that the cliff is flat, i.e. the car came off the cliff horizontally). Question: v = ? [m/s] Given: h = ∆y = 72 m d = ∆x = 22 m g = 10.0 m/s2 Equation: h = viyt + 12 gt 2 and d = vixt Plug n’ Chug: Step 1: Calculate the time required for the car to freefall from a height of 72 m. h = viyt + 21 gt 2 but since viy = 0, the equation simplifies to h = 12 gt 2 rearranging for the unknown variable, t, yields s s 2h 2(72 m) t= = = 3.79 s g 10.0 m/s2 Step 2: Solve for initial velocity: vix =
d t
=
22 m 3.79 s
= 5.80 m/s
Answer: 5.80 m/s
Example 2
Question: A ball of mass m is moving horizontally with a speed of vi off a cliff of height h. How much time does it take the ball to travel from the edge of the cliff to the ground? Express your answer in terms of g (acceleration due to gravity) and h (height of the cliff). Solution: Since we are solving or how long it takes for the ball to reach ground, any motion in the x direction is not pertinent. To make this problem a little simpler, we will define down as the positive direction and the top of the cliff to be y=0 . In this solution we will use the equation 1 y(t) = yo + voyt + gt 2 2 . 16
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1 y(t) = yo + voyt + gt 2 2 1 h = yo + voyt + gt 2 2 1 2 h = 0 + voy + gt 2 1 2 h = 0 + 0 + gt 2 1 2 h = gt 2s t=
2h g
Chapter 1. Two-Dimensional and Projectile Motion [WEEK 2]
start with the equation substitute h for y(t) because that’s the position of the ball when it hits the ground after time t substitute 0 for yo because the ball starts at the top of the cliff substitute 0 for voy becauese the ball starts with no vertical component to it’s velocity simplify the equation solve for t
Watch this Explanation
MEDIA Click image to the left for more content.
Time for Practice
1. A stone is thrown horizontally at a speed of 8.0 m/s from the edge of a cliff 80 m in height. How far from the base of the cliff will the stone strike the ground? 2. A toy truck moves off the edge of a table that is 1.25 m high and lands 0.40 m from the base of the table. a. How much time passed between the moment the car left the table and the moment it hit the floor? b. What was the horizontal velocity of the car when it hit the ground? 3. A hawk in level flight 135 m above the ground drops the fish it caught. If the hawk’s horizontal speed is 20.0 m/s, how far ahead of the drop point will the fish land? 4. A pistol is fired horizontally toward a target 120 m away, but at the same height. The bullet’s velocity is 200 m/s. How long does it take the bullet to get to the target? How far below the target does the bullet hit? 5. A bird, traveling at 20 m/s, wants to hit a waiter 10 m below with his dropping (see image). In order to hit the waiter, the bird must release his dropping some distance before he is directly overhead. What is this distance? 17
1.4. Projectile Motion Problem Solving
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6. Joe Nedney of the San Francisco 49ers kicked a field goal with an initial velocity of 20 m/s at an angle of 60◦ . a. How long is the ball in the air? Hint: you may assume that the ball lands at same height as it starts at. b. What are the range and maximum height of the ball? 7. A racquetball thrown from the ground at an angle of 45◦ and with a speed of 22.5 m/s lands exactly 2.5 s later on the top of a nearby building. Calculate the horizontal distance it traveled and the height of the building. 8. Donovan McNabb throws a football. He throws it with an initial velocity of 30 m/s at an angle of 25◦ . How much time passes until the ball travels 35 m horizontally? What is the height of the ball after 0.5 seconds? (Assume that, when thrown, the ball is 2 m above the ground.) 9. Pablo Sandoval throws a baseball with a horizontal component of velocity of 25 m/s. After 2 seconds, the ball is 40 m above the release point. Calculate the horizontal distance it has traveled by this time, its initial vertical component of velocity, and its initial angle of projection. Also, is the ball on the way up or the way down at this moment in time? 10. Barry Bonds hits a 125 m(4500 ) home run that lands in the stands at an altitude 30 m above its starting altitude. Assuming that the ball left the bat at an angle of 45◦ from the horizontal, calculate how long the ball was in the air. 11. A golfer can drive a ball with an initial speed of 40.0 m/s. If the tee and the green are separated by 100 m, but are on the same level, at what angle should the ball be driven? ( Hint: you should use 2 cos (x) sin (x) = sin (2x) at some point.) 12. How long will it take a bullet fired from a cliff at an initial velocity of 700 m/s, at an angle 30◦ below the horizontal, to reach the ground 200 m below? 13. A diver in Hawaii is jumping off a cliff 45 m high, but she notices that there is an outcropping of rocks 7 m out at the base. So, she must clear a horizontal distance of 7 m during the dive in order to survive. Assuming the diver jumps horizontally, what is his/her minimum push-off speed? 14. If Monte Ellis can jump 1.0 m high on Earth, how high can he jump on the moon assuming same initial velocity that he had on Earth (where gravity is 1/6 that of Earth’s gravity)? 15. James Bond is trying to jump from a helicopter into a speeding Corvette to capture the bad guy. The car is going 30.0 m/s and the helicopter is flying completely horizontally at 100 m/s. The helicopter is 120 m above the car and 440 m behind the car. How long must James Bond wait to jump in order to safely make it into the car?
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Chapter 1. Two-Dimensional and Projectile Motion [WEEK 2]
16. A field goal kicker lines up to kick a 44 yard (40 m) field goal. He kicks it with an initial velocity of 22 m/s at an angle of 55◦ . The field goal posts are 3 meters high.
a. Does he make the field goal? b. What is the ball’s velocity and direction of motion just as it reaches the field goal post (i.e., after it has traveled 40 m in the horizontal direction)? 17. In a football game a punter kicks the ball a horizontal distance of 43 yards (39 m). On TV, they track the hang time, which reads 3.9 seconds. From this information, calculate the angle and speed at which the ball was kicked. (Note for non-football watchers: the projectile starts and lands at the same height. It goes 43 yards horizontally in a time of 3.9 seconds) Answers to Selected Problems
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.
32 m a. 0.5 s b. 0.8 m/s 104 m t = 0.60 s, 1.8 m below target 28 m. a. 3.5 s. b. 35 m; 15 m 40 m; 8.5 m 1.3 seconds, 7.1 meters 50 m; v0y = 30 m/s; 500 ; on the way up 4.4 s 19◦ 0.5 s 2.3 m/s 6m 1.4 seconds a. yes b. 14 m/s @ 23 degrees from horizontal 22 m/s @ 62 degrees
Summary Breaking vectors into components, relative velocity and projectile motion are all covered in this chapter. 19
1.5. References
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1.5 References 1. 2. 3. 4. 5. 6.
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CK-12 Foundation. CK-12 Foundation. CK-12 Foundation. CK-12 Foundation. CK-12 Foundation. CK-12 Foundation.
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