An object thrown upward follows a parabolic height, because gravity applies a constant downward acceleration. The vertex gives the greatest height and when it occurs; the positive root gives the landing time; the y-intercept is the starting height. Context limits the domain.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Try it first
You have 20 meters of fence to go all the way around a rectangle. Try widths of 2, 4, 5, 6 and 8 meters. Which width gives the most area, in meters?
meters
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
What the roots mean
h = −5t² + 20t = −5t(t − 4): roots t = 0 and t = 4. The first is the launch, the second the landing. A negative root, in another model, is mathematically valid and physically meaningless.
From a height
h = −5t² + 20t + 25 starts at 25 m. Landing: −5t² + 20t + 25 = 0, t² − 4t − 5 = 0, (t − 5)(t + 1) = 0, t = 5. The peak is still at t = 2, now 45 m.
Area problems
40 m of fence for a rectangle: w and 20 − w. A = w(20 − w) = −w² + 20w. Vertex at w = 10: a 10 by 10 square, area 100 m². A fixed perimeter with a variable side gives a quadratic area.
Against a wall
Three sides fenced, the wall as the fourth: w + 2h = 40, w = 40 − 2h. A = h(40 − 2h) = −2h² + 40h. Vertex at h = 10, w = 20: area 200 m², twice the enclosed square.
Context limits the domain
Negative time and negative lengths do not exist. For the ball, 0 ≤ t ≤ 4. For the fence, 0 < w < 20. Say which part of the parabola actually applies; the rest is algebra, not the situation.
Ask what the question wants
A maximum needs the vertex; a landing time needs a root; a height at a given time needs substitution; a time at a given height needs solving. Different points on the same parabola.
Step 2: Try It Yourself
Tap and try it out.
A rocket follows h = −5t² + 30t + 35. Find the starting height, the peak, when it lands, and when it is at 60 m
- 1h(0) = 35 mthe starting height: c
Step 0 of 4
Step 3: In Real Life
A fenced garden
With 40 meters of fence against a wall, the area is w(40 − 2w). The vertex gives the best width, 10 meters, and the largest garden, 200 square meters.
Step 4: Watch an Example
One step at a time.
Watch Rosa Read a Roof Toss
A ball is tossed upward from a roof. Its height is h = −5t² + 10t + 15 meters, t seconds after the toss. Rosa finds the start, the peak and the landing.
- Step 1
At t = 0 every term with a t in it is zero, so h = 15. The ball starts 15 m up, at roof height. In any height model, the start is the constant c.
Watch Leo Plan a Pen
Leo has 24 meters of fence for a rectangular pen against a barn wall. The wall is one long side, so the fence makes the other three: two ends, each d meters deep, and a front. Leo sets up the model, and you finish it.
- Step 1
The two ends use 2d meters of fence, so the front gets what is left: 24 − 2d meters. A depth of 4 m, for instance, would leave a front of 16 m.
Step 5: Your Turn
Practice makes it stick.
The Diving Board
Problem 1 of 15
A diver jumps from a board 10 meters above the water. Her height above the water is h = −5t² + 5t + 10 meters, t seconds after she jumps. After how many seconds does she reach the water?
seconds
The Flower Bed
Problem 2 of 15
Mia has 36 feet of edging to go all the way around a rectangular flower bed. She wants the bed to cover 80 square feet. How long is the longer side, in feet?
feet
Read the Model
Decide first what the question wants: a height at a time, a peak, a landing, or a time at a height. Then compute.
Problem 3 of 15
A ball’s height is h = −5t² + 25t meters after t seconds. How high is it at t = 1?
meters
Problem 4 of 15
A ball’s height is h = −5t² + 30t meters after t seconds. What is its greatest height, in meters?
meters
Problem 5 of 15
A ball’s height is h = −5t² + 15t + 20 meters after t seconds. When does it land, in seconds?
seconds
Problem 6 of 15
28 meters of fence go all the way around a rectangle. What is the largest area it can enclose, in square meters?
square meters
Problem 7 of 15
A ball’s height is h = −5t² + 20t + 5 meters after t seconds. At what time is it first 20 meters high, in seconds?
seconds
Whose Step?
One line on each page is a slip. Find it, then fix it.
Problem 8 of 15
Dev wants to know when a ball with height h = −5t² + 20t + 60 meters lands. One line is a slip. Which?
Dev’s landing time for h = −5t² + 20t + 60
- 1h = −5t² + 20t + 60
- 2t = −20/(2 × −5) = 2
- 3h = −5(4) + 20(2) + 60 = 80
- Answerlands at t = 2 s
Problem 9 of 15
When does Dev’s ball land, in seconds?
seconds
Problem 10 of 15
Sam has 32 meters of fence for a rectangular pen against a wall. The wall is one long side, so the fence makes two ends of depth d and a front of length w. Sam looks for the largest area. One line is a slip. Which?
Sam’s largest pen with 32 m of fence against a wall
- 1w + 2d = 32
- 2A = d(32 − d)
- 3A = −d² + 32d
- 4d = −32/(2 × −1) = 16
- Answerlargest area 16 × 16 = 256 m²
Problem 11 of 15
What is the largest area Sam’s pen can have, in square meters?
square meters
Pick the Move
No computing needed. Decide which point of the parabola the question is about.
Problem 12 of 15
A stone is thrown from a cliff. Its height above the sea is h = −5t² + 10t + 40 meters after t seconds. The question: how long until it hits the sea? Which move answers it?
Problem 13 of 15
A model rocket’s height is h = −5t² + 40t + 5 meters after t seconds. The question: how high is it 3 seconds after launch? Which move answers it?
Same Peak, Same Height?
Judge from the shape of the models. No full working needed.
Problem 14 of 15
Without finding any heights, select every model whose peak comes at t = 2 seconds.
Problem 15 of 15
A ball’s height is h = −5t² + 30t, with its peak at t = 3 seconds. Without computing any heights, select every pair of times at which the ball is at the same height.
Step 6: Quick Check
Show what you know.
Question 1 of 2
h = −5t² + 40t. At what time is the height greatest, in seconds?
Question 2 of 2
What does the vertex give in a projectile model?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
A ball is thrown straight up from the ground, so its height is h = −5t² + bt meters after t seconds. For which b, the launch speed in meters per second, does it land exactly 6 seconds after the throw?
meters per second
Stretch 2 of 2
Leo’s 24 meters of fence against a wall made the best pen with a depth of 6 m and a front of 12 m. With 40 meters, the theory’s pen was best at a depth of 10 m and a front of 20 m. How many meters of fence, used the same way, make a largest area of exactly 50 square meters?
meters
What You Learned
- Projectile height and fixed-perimeter area are both quadratic.
- The vertex gives a maximum or minimum; the roots give where the value is zero; the intercept is the start.
- Context restricts which part of the parabola applies.
- Ask what the question wants before computing: vertex, root, or a substitution.
For the grown-up
The misconception this lesson targets is reaching for the vertex whatever the question asks, so the peak time is given as the landing time, or the peak height as the height at some other moment. One parabola answers four different questions at four different points: the start (t = 0), the peak (the vertex), the landing (a root) and a height at a time (substitution). Its cousins are keeping the negative root, which in a story is a time before the throw, and setting up a pen against a wall with the front as 40 − w instead of 40 − 2w, which forgets that two ends are fenced. Asking "what point of the graph is this?" before computing catches all three.