Math II · F-LE.6

Applying Quadratic Functions to Physical Situations Such as Projectile Motion

This objective teaches students that a parabola is not just a graph shape. It is a model for motion, height, area, optimization, and any situation where change itself is changing at a constant rate.

Concept Functions
Domain Linear, Quadratic, and Exponential Models
Read time 9 minutes

What this learning objective is really asking you to learn

This objective asks students to use quadratic functions as models for real physical behavior. A quadratic function has the general form \(f(x)=ax^2+bx+c\), where the squared term is the signature feature. The square does something important: it makes the graph curve. A line changes by the same amount over equal input intervals, but a quadratic does not. Its rate of change changes steadily. That is why quadratics are so useful for modeling situations involving acceleration, curved paths, area, and optimization.

The most famous physical example is projectile motion under gravity. If a ball is thrown upward, it rises quickly at first, slows down, reaches a highest point, and then falls back down faster and faster. A linear function cannot describe that. A line would say the height changes by a constant number of feet or meters each second. But a thrown object does not keep rising at the same speed. Gravity pulls it downward the whole time. Its upward velocity decreases, becomes zero at the top, then turns downward. The path of height over time is a parabola.

A common model in feet is \(h(t)=-16t^2+v_{0t}+h_{0}\), where \(t\) is time in seconds, \(h(t)\) is height in feet, \(v_{0}\) is the initial vertical velocity in feet per second, and \(h_{0}\) is the initial height. The coefficient -16 appears because gravity near Earth's surface accelerates falling objects downward at about \(32 ft/s^2\), and the position formula uses half the acceleration. In meters, a common model is \(h(t)=-4.9t^2+v_{0t}+h_{0}\), because gravitational acceleration is about \(9.8 m/s^2\). Students do not need to become physicists to understand the math, but they do need to see what each part of the equation represents.

The constant term represents the height at time zero. If a ball is released from a platform 6 feet above the ground, then \(h_{0}=6\). The linear term represents the initial upward or downward velocity. If the coefficient of \(t\) is positive, the object starts by moving upward. If it is negative, the object is launched downward or already falling. The quadratic coefficient represents the acceleration effect, and for ordinary vertical projectile motion on Earth it is negative because gravity pulls downward. The parabola opens downward, so the graph has a maximum instead of a minimum.

This standard also asks students to interpret the key features of the graph. The y-intercept is the starting height. The vertex is the maximum height and the time when that maximum occurs. The positive x-intercept, when it exists, can represent the time when the object reaches the ground. The domain is usually restricted to meaningful time values, because negative time may not belong to the story and times after the object hits the ground may not belong to the flight. The range is restricted by the possible heights during the modeled event.

Students are not just plugging into formulas. They are translating between a story, an equation, a graph, and a set of decisions. That translation is the heart of mathematical modeling.

Why students should learn this math

Students should learn this math because quadratics are one of the first places where school algebra starts to look like the physical world. When students ask, “Why do I need this?” projectile motion gives a direct answer: because the world is full of curved motion, and quadratics are the first practical tool for describing it.

A basketball shot is not a straight line from the player's hand to the hoop. It follows an arc. A soccer ball, a thrown key, a water fountain, a diving athlete, a tossed phone, a firework shell, a drone's vertical motion, and a jumping skateboarder all involve curved height-over-time behavior. Even when air resistance matters and the exact path is not a perfect parabola, the quadratic model is a powerful first approximation. It teaches students how to identify the important quantities: where something starts, how fast it starts moving, what force changes that motion, when it reaches a peak, and when it comes back down.

This is not only about sports or stunts. Engineers use versions of this thinking in designing ramps, projectiles, robotics motions, vehicle safety systems, amusement rides, drainage paths, and mechanical arms. Video game developers use parabolic arcs to make jumps, thrown objects, and falling objects feel believable. Animators use curved motion so movement looks natural instead of robotic. Architects and designers use parabolic shapes in arches, reflectors, suspension forms, and aesthetic structures. Data analysts use quadratic models when change accelerates or decelerates in a roughly constant way. A student who understands quadratics has a mental tool for recognizing “the rate itself is changing.”

There is also a deeper life skill here: distinguishing a model from reality. A quadratic projectile model is useful, but it has assumptions. It usually assumes constant gravity, no air resistance, motion close to Earth, and a simple vertical dimension. If a feather falls, air resistance dominates. If a rocket fires engines, acceleration is not just gravity. If a baseball spins, lift and drag can curve the path differently. Learning quadratics helps students learn how models work: not as magic truth, but as simplified maps of reality. A good model is not judged by whether it includes everything. It is judged by whether it captures the important behavior for the question being asked.

This objective also gives meaning to many algebra skills students previously learned. Factoring is not just a symbolic trick; it can find when height equals zero. Completing the square is not just a procedure; it can reveal maximum height and the time it occurs. The quadratic formula is not just a memorized expression; it can find landing time when factoring is inconvenient. Graphing is not just plotting points; it can show the whole story of an object's motion. Domain restrictions are not artificial; they keep the answer inside the real event.

For students who struggle with math because it feels disconnected, this objective can be a turning point. It says: the equation is a machine for telling a story about motion.

The historical machinery behind quadratic motion

Quadratic relationships have roots in ancient geometry, but their connection to motion became especially powerful during the scientific revolution. Ancient mathematicians studied parabolas as conic sections. A parabola can be formed by slicing a cone at a particular angle. Greek geometers such as Apollonius studied conic sections long before modern algebraic notation existed. At that time, parabolas were geometric objects, not necessarily function graphs on coordinate axes.

The major shift came when algebra, coordinate geometry, and physics began to merge. René Descartes and Pierre de Fermat helped develop analytic geometry, where geometric shapes could be represented by equations. A parabola could now be described with symbols, not only drawn with compass and straightedge. This opened the door to using algebraic curves as models of real phenomena.

Galileo Galilei's work on falling bodies and projectile motion was a landmark. He studied how distance fallen under constant acceleration relates to time and showed that uniform acceleration leads to a square relationship. If speed increases at a constant rate, then distance grows like time squared. This is the heart of why projectile height uses a quadratic term. The object is not merely moving; its velocity is changing.

Later, Isaac Newton built a broader mathematical framework for motion through laws of motion and gravitation. In Newtonian mechanics, position, velocity, and acceleration are connected. In the language students will later see in calculus, velocity is the rate of change of position, and acceleration is the rate of change of velocity. If acceleration is constant, position is quadratic in time. Students in Integrated Math II are not required to use calculus, but they are meeting the algebraic shadow of calculus. The quadratic model is what constant acceleration looks like before formal derivatives and integrals enter the picture.

The historical importance is huge. The same mathematical idea that helps a student model a tossed ball also helped humanity understand falling bodies, artillery paths, planetary approximations, engineering design, and eventually modern physics education. Quadratic motion is part of the machinery that made mathematical science possible.

The technical machinery: how the model works

A projectile-height model usually has the form \(h(t)=at^2+bt+c\), with \(a<0\) for ordinary vertical motion under gravity. The input is time. The output is height. The coefficient \(a\) controls the curvature and is tied to acceleration. The coefficient \(b\) represents initial velocity. The constant \(c\) represents initial height.

Suppose a ball is thrown upward from a height of 5 feet with an initial vertical velocity of 48 feet per second. A model is \(h(t)=-16t^2+48t+5\). At \(t=0\), the height is 5, so the y-intercept tells us the starting height. The coefficient 48 tells us the ball starts by moving upward. The coefficient -16 tells us gravity curves the path downward.

To find the maximum height, students can use the vertex. For \(h(t)=at^2+bt+c\), the time coordinate of the vertex is \(t=-b/(2a)\). In this example, \(t=-48/(2(-16))=1.5\). The ball reaches its highest point after 1.5 seconds. Then \(h(1.5)=-16(1.5)^2+48(1.5)+5=-36+72+5=41\). The maximum height is 41 feet.

To find when the ball hits the ground, students solve \(-16t^2+48t+5=0\). This may require the quadratic formula because the numbers do not factor nicely. The positive solution is the meaningful landing time. The negative solution, if present, is usually not meaningful for the modeled flight because it would describe a time before the launch in the mathematical extension of the parabola.

That last idea is crucial. The graph may extend forever, but the situation does not. The mathematical function has a full domain of real numbers, but the physical model has a restricted domain. For this ball, the story begins at \(t=0\) and ends when the ball hits the ground. A negative time or a time after ground impact may not answer the original question.

Students should also learn to connect different forms of a quadratic to different information. Standard form \(at^2+bt+c\) makes the starting height easy to see. Vertex form \(a(t-h)^2+k\) makes maximum or minimum height easy to see. Factored form \(a(t-r_{1})(t-r_{2})\) makes zeros easy to see. No form is “the best” in every situation. The best form depends on the question.

What can go wrong, and how to fix it

One common mistake is treating every x-intercept as physically meaningful. In a projectile problem, time cannot be negative once the launch moment has been chosen as \(t=0\). A negative solution may be mathematically valid for the extended parabola but irrelevant to the event.

Another mistake is forgetting units. If height is in feet and time is in seconds, the gravity coefficient is usually -16, not -4.9. If height is in meters, the model usually uses -4.9. Mixing units can produce nonsense answers. Units are not decoration; they are part of the model.

A third mistake is assuming the vertex always gives the answer. The vertex gives maximum or minimum output. If the question asks when the object hits the ground, students need an x-intercept. If the question asks the starting height, they need the y-intercept. If the question asks height after two seconds, they evaluate the function. Matching the method to the question is part of mastery.

A fourth mistake is believing the model is exact in every physical situation. Air resistance, wind, spin, changing gravitational fields, engines, collisions, and measurement error can all affect motion. Quadratic projectile motion is a powerful model, but it is a model. Students should learn both to use it and to understand its assumptions.

Where this fits into the big map of math

This objective sits at the crossroads of algebra, functions, physics, and modeling. It uses quadratics from algebra, graph features from functions, units from number and quantity, and physical interpretation from science. It prepares students for calculus because constant acceleration produces quadratic position, linear velocity, and constant acceleration. Later, derivatives will formalize those relationships.

It also prepares students for engineering and data modeling. Any time a process has a changing rate that changes at a roughly constant amount, quadratic thinking may apply. In optimization, a quadratic can describe profit, area, cost, or error. In geometry, area often produces quadratic expressions. In physics, constant acceleration produces quadratic position. In statistics and machine learning, squared error becomes a major idea. This single graph family appears in many branches of the mathematical map.

Mastery means students can look at a parabola and see a story. They can look at an equation and identify initial value, velocity, curvature, peak, roots, and meaningful domain. They can explain why a ball slows down as it rises and speeds up as it falls. They can decide whether an answer makes sense. That is not just algebra. That is mathematical literacy about motion.

Problem Library

Problems in the App From This Objective

180 problems across 15 archetypes in the app.

interpret y-intercept.
12 problems Warmup Practice Mixed Review Assessment
Problem 1

Identify the initial height in projectile function h(t)=-16t^2+48t+5.

Problem 2

Identify the initial height in projectile function h(t)=-4.9t^2+20t+1.5.

Problem 3

Identify the initial height in projectile function h(t)=-16t^2+64t.

Problem 4

Identify the initial height in projectile function h(t)=-16(t-2)^2+80.

Problem 5

Identify the initial height in projectile function h(t)=-16t^2+32t+10.

Problem 6

Identify the initial height in projectile function h(t)=-4.9t^2+10t+2.5.

Open in simulator
Problem 7

Identify the initial height in projectile function h(t)=-16t^2+100t.

Problem 8

Identify the initial height in projectile function h(t)=-16(t-3)^2+144.

Problem 9

Identify the initial height in projectile function h(t)=-4.9(t-1)^2+10.

Problem 10

Identify the initial height in projectile function h(t)=-16t^2+128t+20.

Problem 11

Identify the initial height in projectile function h(t)=-16(t-1.5)^2+100.

Problem 12

Identify the initial height in projectile function h(t)=-4.9t^2+30t+0.8.

interpret linear coefficient in context.
12 problems Warmup Practice Mixed Review Assessment
Problem 13

Identify the initial velocity in projectile function h(t)=-16t^2+48t+5.

Problem 14

Identify the initial velocity in projectile function h(t)=-4.9t^2-12t+20.

Problem 15

Identify the initial velocity in projectile function h(t)=-16t^2+64t.

Problem 16

Identify the initial velocity in projectile function h(t)=-16t^2+0t+80.

Problem 17

Identify the initial velocity in projectile function h(t)=-16t^2+32t+10.

Problem 18

Identify the initial velocity in projectile function h(t)=-16t^2-24t+100.

Problem 19

Identify the initial velocity in projectile function h(t)=-4.9t^2+9.8t+1.

Problem 20

Identify the initial velocity in projectile function h(t)=-4.9t^2-5t+15.

Problem 21

Identify the initial velocity in projectile function h(t)=-4.9t^2+0t+50.

Open in simulator
Problem 22

Identify the initial velocity in projectile function h(t)=-16t^2+80t.

Problem 23

Identify the initial velocity in projectile function h(t)=-16t^2-10t+25.

Problem 24

Identify the initial velocity in projectile function h(t)=-4.9t^2+20t+0.

find vertex and interpret y-value.
12 problems Warmup Practice Mixed Review Assessment
Problem 25

Find maximum height of projectile h(t)=-16t^2+64t+5.

Problem 26

Find maximum height of projectile h(t)=-16t^2+32t+48.

Problem 27

Find maximum height of projectile h(t)=-4.9t^2+19.6t+2.

Problem 28

Find maximum height of projectile h(t)=-16(t-3)^2+100.

Problem 29

Find maximum height of projectile h(t)=-16t^2+80t+10.

Problem 30

Find maximum height of projectile h(t)=-16t^2+96t.

Problem 31

Find maximum height of projectile h(t)=-4.9t^2+29.4t+5.

Open in simulator
Problem 32

Find maximum height of projectile h(t)=-4.9t^2+9.8t+10.

Problem 33

Find maximum height of projectile h(t)=-16(t-2)^2+80.

Problem 34

Find maximum height of projectile h(t)=-16(t-4)^2+120.

Problem 35

Find maximum height of projectile h(t)=-4.9(t-1.5)^2+25.

Problem 36

Find maximum height of projectile h(t)=-4.9(t-2.5)^2+30.5.

identify vertex x-value.
12 problems Warmup Practice Mixed Review Assessment
Problem 37

Find when projectile h(t)=-16t^2+64t+5 reaches maximum height.

Problem 38

Find when projectile h(t)=-16t^2+32t+48 reaches maximum height.

Problem 39

Find when projectile h(t)=-4.9t^2+19.6t+2 reaches maximum height.

Problem 40

Find when projectile h(t)=-16(t-3)^2+100 reaches maximum height.

Open in simulator
Problem 41

Find when projectile h(t)=-5t^2+30t+10 reaches maximum height.

Problem 42

Find when projectile h(t)=-9.8t^2+98t+1 reaches maximum height.

Problem 43

Find when projectile h(t)=-2t^2+16t-3 reaches maximum height.

Problem 44

Find when projectile h(t)=-0.5(t-7)^2+20 reaches maximum height.

Problem 45

Find when projectile h(t)=-16t^2+128t+200 reaches maximum height.

Problem 46

Find when projectile h(t)=-10t^2+50t reaches maximum height.

Problem 47

Find when projectile h(t)=-4(t-0.5)^2+10 reaches maximum height.

Problem 48

Find when projectile h(t)=-3t^2+18t+7 reaches maximum height.

solve quadratic for height zero and select valid time.
12 problems Warmup Practice Mixed Review Assessment
Problem 49

Find when projectile h(t)=-16t^2+64t reaches the ground.

Problem 50

Find when projectile h(t)=-16t^2+32t+48 reaches the ground.

Problem 51

Find when projectile h(t)=-4.9t^2+19.6t reaches the ground.

Open in simulator
Problem 52

Find when projectile h(t)=-16(t-2)^2+64 reaches the ground.

Problem 53

Find when projectile h(t)=-16t^2+80t reaches the ground.

Problem 54

Find when projectile h(t)=-4.9t^2+29.4t reaches the ground.

Problem 55

Find when projectile h(t)=-16t^2+16t+32 reaches the ground.

Problem 56

Find when projectile h(t)=-16t^2+8t+24 reaches the ground.

Problem 57

Find when projectile h(t)=-16(t-3)^2+144 reaches the ground.

Problem 58

Find when projectile h(t)=-4.9(t-5)^2+122.5 reaches the ground.

Problem 59

Find when projectile h(t)=-16t^2+40t+56 reaches the ground.

Problem 60

Find when projectile h(t)=-16t^2+20t+6 reaches the ground.

solve quadratic equation `h(t)=target`.
12 problems Warmup Practice Mixed Review Assessment
Problem 61

Find when projectile h(t)=-16t^2+64t reaches target height 48.

Problem 62

Find when projectile h(t)=-16t^2+32t+48 reaches target height 48.

Problem 63

Find when projectile h(t)=-16(t-2)^2+64 reaches target height 64.

Problem 64

Find when projectile h(t)=-16t^2+16t+20 reaches target height 40.

Problem 65

Find when projectile h(t)=-16t^2+80t reaches target height 64.

Problem 66

Find when projectile h(t)=-16t^2+96t reaches target height 80.

Problem 67

Find when projectile h(t)=-16t^2+64t+32 reaches target height 80.

Problem 68

Find when projectile h(t)=-16t^2+48t+64 reaches target height 64.

Open in simulator
Problem 69

Find when projectile h(t)=-16t^2+112t reaches target height 96.

Problem 70

Find when projectile h(t)=-16t^2+64t+16 reaches target height 64.

Problem 71

Find when projectile h(t)=-16t^2+32t+96 reaches target height 96.

Problem 72

Find when projectile h(t)=-16t^2+128t reaches target height 112.

connect symmetry around maximum.
12 problems Warmup Practice Mixed Review Assessment
Problem 73

Interpret two times 1, 3 with the same projectile height in model h(t)=-16(t-2)^2+64.

Problem 74

Interpret two times 0.5, 1.5 with the same projectile height in model h(t)=-16(t-1)^2+40.

Problem 75

Interpret two times 2, 4 with the same projectile height in model h(t)=-4.9(t-3)^2+50.

Problem 76

Interpret two times 1, 3 with the same projectile height in model h(t)=-16t(t-4).

Problem 77

Interpret two times 4, 6 with the same projectile height in model h(t)=-9.8(t-5)^2+100.

Problem 78

Interpret two times 2, 3 with the same projectile height in model h(t)=-16(t-2.5)^2+80.

Problem 79

Interpret two times 3, 7 with the same projectile height in model h(t)=-4.9t(t-10).

Open in simulator
Problem 80

Interpret two times 0.5, 3.5 with the same projectile height in model h(t)=-16t^2+64t.

Problem 81

Interpret two times 1, 7 with the same projectile height in model h(t)=-9.8(t-4)^2+70.

Problem 82

Interpret two times 2, 4 with the same projectile height in model h(t)=-16(t-0.5)(t-5.5).

Problem 83

Interpret two times 1, 2 with the same projectile height in model h(t)=-4.9(t-1.5)^2+20.

Problem 84

Interpret two times 2.5, 3.5 with the same projectile height in model h(t)=-4.9t^2+29.4t.

restrict time from launch to landing or observation interval.
12 problems Warmup Practice Mixed Review Assessment
Problem 85

Determine the meaningful domain for projectile model h(t)=-16t^2+64t in context from launch until landing.

Problem 86

Determine the meaningful domain for projectile model h(t)=-16t^2+32t+48 in context while at or above ground.

Problem 87

Determine the meaningful domain for projectile model h(t)=-4.9t^2+19.6t in context from launch to ground.

Problem 88

Determine the meaningful domain for projectile model h(t)=-16(t-2)^2+64 in context visible flight above ground.

Problem 89

Determine the meaningful domain for projectile model h(t)=-16t^2+80t in context from launch until it hits the ground.

Problem 90

Determine the meaningful domain for projectile model h(t)=-16t^2+48t+160 in context while the object is above the ground.

Problem 91

Determine the meaningful domain for projectile model h(t)=-4.9t^2+29.4t in context from its initial launch until it lands.

Problem 92

Determine the meaningful domain for projectile model h(t)=-4.9t^2+9.8t+14.7 in context the time the projectile is in the air.

Open in simulator
Problem 93

Determine the meaningful domain for projectile model h(t)=-16t^2+96t in context from the moment it is launched until it returns to the ground.

Problem 94

Determine the meaningful domain for projectile model h(t)=-16t^2+64t+80 in context for the duration of its flight above ground.

Problem 95

Determine the meaningful domain for projectile model h(t)=-4.9t^2+39.2t in context from the point of projection until it lands.

Problem 96

Determine the meaningful domain for projectile model h(t)=-4.9t^2+14.7t+49 in context while the object is in the air.

use minimum/maximum over valid domain.
12 problems Warmup Practice Mixed Review Assessment
Problem 97

Determine the meaningful range for projectile model h(t)=-16t^2+64t over domain 0 <= t <= 4.

Open in simulator
Problem 98

Determine the meaningful range for projectile model h(t)=-16t^2+32t+48 over domain 0 <= t <= 3.

Problem 99

Determine the meaningful range for projectile model h(t)=-16(t-2)^2+64 over domain 0 <= t <= 4.

Problem 100

Determine the meaningful range for projectile model h(t)=-4.9(t-2)^2+21.6 over domain 0 <= t <= 4.

Problem 101

Determine the meaningful range for projectile model h(t)=-16t^2+80t over domain 0 <= t <= 5.

Problem 102

Determine the meaningful range for projectile model h(t)=-4.9t^2+19.6t+5 over domain 0 <= t <= 4.24.

Problem 103

Determine the meaningful range for projectile model h(t)=-16(t-1)^2+50 over domain 0 <= t <= 2.

Problem 104

Determine the meaningful range for projectile model h(t)=-16t^2+96t+10 over domain 4 <= t <= 6.

Problem 105

Determine the meaningful range for projectile model h(t)=-16t^2+96t+10 over domain 0 <= t <= 2.

Problem 106

Determine the meaningful range for projectile model h(t)=-4.9t^2+9.8t+10 over domain 0 <= t <= 2.74.

Problem 107

Determine the meaningful range for projectile model h(t)=-16(t-3)^2+144 over domain 0 <= t <= 4.

Problem 108

Determine the meaningful range for projectile model h(t)=-0.5t^2+5t+12 over domain 0 <= t <= 12.

place intercepts, vertex, and domain.
12 problems Warmup Practice Mixed Review Assessment
Problem 109

List sketch features for projectile model h(t)=-16t(t-4).

Problem 110

List sketch features for projectile model h(t)=-16t^2+32t+48.

Problem 111

List sketch features for projectile model h(t)=-16(t-2)^2+64.

Problem 112

List sketch features for projectile model h(t)=-4.9t(t-4).

Problem 113

List sketch features for projectile model h(t)=-16t(t-5).

Problem 114

List sketch features for projectile model h(t)=-16t^2+64t+80.

Open in simulator
Problem 115

List sketch features for projectile model h(t)=-16(t-3)^2+144.

Problem 116

List sketch features for projectile model h(t)=-16t^2+48t+64.

Problem 117

List sketch features for projectile model h(t)=-4.9t(t-6).

Problem 118

List sketch features for projectile model h(t)=-16t^2+80t+96.

Problem 119

List sketch features for projectile model h(t)=-16(t-1.5)^2+100.

Problem 120

List sketch features for projectile model h(t)=-4.9t(t-5).

compare initial height, velocity, maximum, and landing time.
12 problems Warmup Practice Mixed Review Assessment
Problem 121

Compare projectile models A(t)=-16t^2+64t and B(t)=-16t^2+48t+20.

Problem 122

Compare projectile models A(t)=-16(t-2)^2+64 and B(t)=-16(t-3)^2+80.

Problem 123

Compare projectile models A starts at 0 and lands at 4 and B starts at 20 and lands at 5.

Problem 124

Compare projectile models A(t)=-4.9t^2+19.6t and B(t)=-4.9t^2+9.8t+10.

Problem 125

Compare projectile models A(t) = -16t^2 + 32t + 10 and B(t) = -16t^2 + 16t + 30.

Open in simulator
Problem 126

Compare projectile models A(t) = -16(t-1)^2 + 26 and B(t) = -16(t-1.5)^2 + 36.

Problem 127

Compare projectile models A starts at 5 and lands at 3 seconds and B starts at 0 and lands at 2 seconds.

Problem 128

Compare projectile models A(t) = -16t^2 + 80t and B(t) = -16t^2 + 64t + 16.

Problem 129

Compare projectile models A(t) = -4.9t^2 + 14.7t + 5 and B(t) = -4.9t^2 + 14.7t.

Problem 130

Compare projectile models A(t) = -16t^2 + 64t and B(t) = -2.7t^2 + 64t.

Problem 131

Compare projectile models A(t) = -16(t-2)^2 + 64 and B(t) = -16(t-1)^2 + 48.

Problem 132

Compare projectile models A reaches a max height of 100 feet at 3 seconds and lands at 6 seconds and B reaches a max height of 80 feet at 2 seconds and lands at 4 seconds.

solve and interpret constraints.
12 problems Warmup Practice Mixed Review Assessment
Problem 133

Use projectile model h(t)=-16(t-2)^2+64 to answer design question Does it clear a 50-foot obstacle at t=2?.

Problem 134

Use projectile model h(t)=-16t^2+64t to answer design question Is the ball above 48 feet at t=1 and t=3?.

Problem 135

Use projectile model h(t)=-16t^2+32t+48 to answer design question Does it stay above ground through t=2?.

Problem 136

Use projectile model h(t)=-4.9t^2+19.6t to answer design question Does it reach 25 meters?.

Problem 137

Use projectile model h(t)=-16t^2+80t+10 to answer design question Does it reach a height of 110 feet?.

Problem 138

Use projectile model h(t)=-4.9t^2+9.8t+5 to answer design question Is the object at least 8 meters high at t=1 second?.

Problem 139

Use projectile model h(t)=-16t^2+64t+80 to answer design question Does it stay in the air for 6 seconds?.

Problem 140

Use projectile model h(t)=-16(t-3)^2+100 to answer design question Is the object above 80 feet at t=4 seconds?.

Problem 141

Use projectile model h(t)=-4.9t(t-10) to answer design question Can it reach a height of 125 meters?.

Problem 142

Use projectile model h(t)=-16t^2+48t+12 to answer design question Is the object still rising at t=1.5 seconds?.

Problem 143

Use projectile model h(t)=-16t^2+96t to answer design question Does it clear a 140-foot building?.

Open in simulator
Problem 144

Use projectile model h(t)=-4.9t^2+29.4t+10 to answer design question Will it be in the air for more than 6 seconds?.

approximate vertex and zeros.
12 problems Warmup Practice Mixed Review Assessment
Problem 145

Estimate projectile features from data table: t 0,1,2,3,4; h 0,48,64,48,0.

Problem 146

Estimate projectile features from data graph shows vertex near (1.5,36) and intercepts near t=0 and t=3.

Problem 147

Estimate projectile features from data table: t 0,1,2,3; h 48,64,48,0.

Problem 148

Estimate projectile features from data graph crosses height 0 at t=0 and t=5 with peak near t=2.5.

Problem 149

Estimate projectile features from data table: t 0,1,2,3,4; h 0,32,48,32,0.

Problem 150

Estimate projectile features from data graph shows vertex at (3, 72) and intercepts at t=0 and t=6.

Problem 151

Estimate projectile features from data table: t 0, 0.5, 1, 1.5, 2; h 0, 16, 20, 16, 0.

Problem 152

Estimate projectile features from data graph shows peak height 90 at t=4.5 and lands at t=9.

Problem 153

Estimate projectile features from data table: t 0, 2, 4, 6, 8; h 0, 60, 80, 60, 0.

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Problem 154

Estimate projectile features from data graph shows a parabolic trajectory with vertex at (2.2, 48) and x-intercepts at (0,0) and (4.4,0).

Problem 155

Estimate projectile features from data table: t 0, 0.25, 0.5, 0.75, 1; h 0, 10, 12, 10, 0.

Problem 156

Estimate projectile features from data graph indicates peak height of 120 feet after 5 seconds, landing at 10 seconds.

recognize assumptions such as no air resistance and domain limits.
12 problems Warmup Practice Mixed Review Assessment
Problem 157

Explain limitation of projectile model h(t)=-16t^2+64t in situation a real baseball hit outdoors.

Problem 158

Explain limitation of projectile model h(t)=-4.9t^2+20t+2 in situation a rocket with engine thrust.

Problem 159

Explain limitation of projectile model h(t)=-16t(t-4) in situation asking for h(10).

Problem 160

Explain limitation of projectile model h(t)=-16t^2+32t+5 in situation measuring very precise motion.

Problem 161

Explain limitation of projectile model h(t) = -4.9t^2 + 10t + 100 in situation a satellite orbiting Earth.

Problem 162

Explain limitation of projectile model h(t) = -16t^2 + 80t in situation calculating the height of a ball at t = -2 seconds.

Problem 163

Explain limitation of projectile model h(t) = -16t^2 + 50t + 3 in situation a bird flying upwards after jumping.

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Problem 164

Explain limitation of projectile model h(t) = -4.9t^2 + 15t in situation a ball thrown on the Moon.

Problem 165

Explain limitation of projectile model h(t) = -16t^2 + 1000t in situation a ballistic missile traveling across continents.

Problem 166

Explain limitation of projectile model h(t) = -4.9t^2 + 30t in situation a feather falling from a height.

Problem 167

Explain limitation of projectile model h(t) = -16t^2 + 60t + 4 in situation a hot air balloon rising steadily.

Problem 168

Explain limitation of projectile model h(t) = -4.9t^2 + 25t in situation a diver entering the water.

catch wrong root, invalid negative time, vertex/intercept confusion, and unit mistakes.
12 problems Warmup Practice Mixed Review Assessment
Problem 169

Correct the projectile-modeling error in For h(t)=-16t^2+64t, landing times are t=0 and t=-4.

Problem 170

Correct the projectile-modeling error in The vertex height of h(t)=-16(t-2)^2+64 is 2 feet.

Problem 171

Correct the projectile-modeling error in Initial height in h(t)=-16t^2+48t+5 is 48.

Problem 172

Correct the projectile-modeling error in A negative root t=-1 is a valid time before launch.

Problem 173

Correct the projectile-modeling error in For h(t)=-16t^2+32t+10, the projectile hits the ground at t=10.

Problem 174

Correct the projectile-modeling error in For h(t)=-16t^2+96t+10, the time to reach maximum height is t=6 seconds.

Problem 175

Correct the projectile-modeling error in For h(t)=-16t^2+64t+5, the maximum height is t=2.

Problem 176

Correct the projectile-modeling error in For h(t)=-16t^2+16t+32, the projectile lands at t=-1 and t=2.

Problem 177

Correct the projectile-modeling error in The maximum height of a projectile is 30 seconds.

Problem 178

Correct the projectile-modeling error in For h(t)=-16t^2+80t, the axis of symmetry t=2.5 means the projectile reaches 2.5 feet high.

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Problem 179

Correct the projectile-modeling error in For h(t)=-16t^2+112t, the initial velocity is 16.

Problem 180

Correct the projectile-modeling error in For h(t)=-16t^2+64t, the projectile lands when t = -b/(2a) = 2.