Quadratic Relations of the Form y = ax² + bx + c · Quadratic Expressions and Equations

Solving Problems: Heights of Thrown Objects

4.2 4.1 3.8About 70 minutes

Learning goals

I can interpret each part of a height model h = −4.9t² + vt + h₀.
I can find the maximum height, the landing time and the starting height of a thrown object.
I can find when an object is at a given height, and the interval during which it is above a given height.
I can build a model from a vertex and a point, and judge whether answers are reasonable.

Words to know

model
An equation or graph that represents a real situation, used to make predictions.

initial height
The height when t = 0. In h = at² + bt + c, it is c.

time interval
A stretch of time between two values, such as 0.05 s < t < 3.95 s.

reasonable
Makes sense in the real situation, for example a positive time or a believable height.

What goes up...

A volleyball serve, a basketball shot, a snowball tossed over a fence, a diver leaving a platform: anything moving freely through the air (ignoring air resistance) follows a parabola. Near Earth's surface its height in metres after t seconds can be modelled by

h = −4.9t² + vt + h₀

where h₀ is the starting height and v is the starting upward speed in metres per second. The −4.9 comes from gravity. (Some textbook problems round it to −5 to keep the arithmetic simple.)

Every question about such a model is one of the skills you already have. The trick is to translate the question into the right mathematical task.

Question in words	What to do mathematically
How high was it at the start?	Find h when t = 0 (the constant term).
What is the maximum height? When does it happen?	Find the vertex: complete the square, average the zeros, use t = −b ÷ (2a), or use technology.
When does it hit the ground?	Solve h = 0 and keep the positive root.
When is it at a height of 3 m?	Solve h = 3 (usually two answers: going up and coming down).
For how long is it above 3 m?	Find both times when h = 3, then the interval between them.
Does it ever reach 30 m?	Compare with the maximum height, or check the discriminant of h = 30.

A translation table for problems about heights of thrown objects.

Worked example

Example 1: A thrown ball

A ball is thrown upward from a height of 2 m. Its height is modelled by h = −5t² + 20t + 2. Find (a) the maximum height and when it occurs, (b) when the ball hits the ground, and (c) the time interval during which the ball is higher than 3 m.

(a) Complete the square: h = −5(t² − 4t) + 2 = −5(t² − 4t + 4 − 4) + 2 = −5(t − 2)² + 20 + 2 = −5(t − 2)² + 22. Maximum height 22 m at t = 2 s.
(b) Solve −5t² + 20t + 2 = 0 with the formula: t = (−20 ± √(400 + 40)) ÷ (−10) = (−20 ± √440) ÷ (−10). The roots are t ≈ −0.10 and t ≈ 4.10. Time must be positive: it lands at about 4.10 s.
(c) Solve h = 3: −5t² + 20t + 2 = 3, so −5t² + 20t − 1 = 0, or 5t² − 20t + 1 = 0.
t = (20 ± √(400 − 20)) ÷ 10 = (20 ± √380) ÷ 10, so t ≈ 0.05 or t ≈ 3.95.
The parabola opens down, so h > 3 between these times: 0.05 s < t < 3.95 s, which is about 3.9 s.

(a) 22 m at 2 s; (b) ≈ 4.10 s; (c) between about 0.05 s and 3.95 s.

h = −5t² + 20t + 2. The dashed line is h = 3. The ball is above 3 m in the shaded time interval, from about 0.05 s to 3.95 s. Notice that the times are symmetric about the vertex time, t = 2.

Worked example

Example 2: A platform dive

A diver leaves a 10 m platform. Her height above the water is modelled by h = −4.9t² + 3t + 10. Find her maximum height and when she enters the water. Round to two decimal places.

Axis: t = −b ÷ (2a) = −3 ÷ (−9.8) ≈ 0.31 s.
Maximum height: substitute t = 3 ÷ 9.8, or use technology: h ≈ 10.46 m.
Enters the water when h = 0: t = (−3 ± √(9 + 196)) ÷ (−9.8) = (−3 ± √205) ÷ (−9.8).
√205 ≈ 14.318, giving t ≈ −1.15 (reject) or t ≈ 1.77.
Reasonableness: she rises only about half a metre, and a little under 2 s from a 10 m platform to the water is believable. ✓

Maximum height ≈ 10.46 m after about 0.31 s; she enters the water after ≈ 1.77 s.

h = −4.9t² + 3t + 10. The graph starts at the platform height (0, 10), rises slightly to the maximum, then falls to the water at about 1.77 s. The model only makes sense from t = 0 until the diver reaches the water.

Worked example

Example 3: Build the model from facts

A soccer ball is kicked from the ground. It reaches a maximum height of 8 m after 1.25 s. Find a model in vertex form, then find when it lands and its height at 2 s.

Vertex (1.25, 8): h = a(t − 1.25)² + 8.
It starts on the ground, so (0, 0) is on the graph: 0 = a(1.5625) + 8, so a = −8 ÷ 1.5625 = −5.12.
Model: h = −5.12(t − 1.25)² + 8.
By symmetry it lands at t = 2 × 1.25 = 2.5 s.
At t = 2: h = −5.12(0.75)² + 8 = −5.12(0.5625) + 8 = −2.88 + 8 = 5.12 m.

h = −5.12(t − 1.25)² + 8; lands at 2.5 s; height at 2 s is 5.12 m.

Solving problems well

Define the variables and units (t in seconds, h in metres).
Translate each question into a task: vertex, zeros, or solving h = a number.
Reject roots outside the situation (negative times, times after landing).
Round only at the end, and answer in a sentence with units.
Ask "Is this reasonable?" A ball thrown by hand does not rise 400 m.

Hands-on activity

Time your own toss

You need: a soft ball (a foam ball or a pair of rolled-up socks), a phone or watch with a stopwatch, a helper, and an open space outdoors or in a large room with nothing breakable. A grown-up should say the space is safe.

Toss the ball straight up and catch it at the same height you released it. Your helper times the flight from release to catch, in seconds (T).
Repeat three times and average your times.
If the ball leaves and returns to your hand, a model for its height above your hand is h = −4.9t(t − T). The maximum is halfway, at t = T ÷ 2.
Calculate the maximum height above your hand: h = 4.9 × (T ÷ 2)².
Does your answer look reasonable compared with what you saw?

Model answer: if T = 2 s, the ball rose 4.9 × 1² = 4.9 m above your hand. If T = 1.2 s, it rose 4.9 × 0.6² ≈ 1.8 m. Timing by hand is not exact, so expect your answers to vary a little.

Just for fun

On the Moon, gravity is about one-sixth as strong as on Earth, so the −4.9 in our models would become about −0.81. A ball tossed at the same speed would stay up about six times longer and rise six times higher. Moon volleyball would be a very slow game.

Your turn

A ball's height is h = −5t² + 30t + 1.5 (h in metres, t in seconds). What was its starting height?

For h = −5t² + 30t + 1.5, find the maximum height and when it occurs.

For h = −5t² + 30t + 1.5, how high is the ball after 1 s?

For h = −5t² + 30t + 1.5, when does the ball hit the ground? (Two decimal places.)

The ball is at 26.5 m at t = 1 s. Without solving, when else is it at 26.5 m? Explain.

For h = −5t² + 30t + 1.5, for how long is the ball higher than 40 m?

More practice: warm-up

In h = −4.9t² + 15t + 2, what does the 2 represent?

State the maximum height for h = −5(t − 2)² + 25, and when it occurs.

For h = −5(t − 2)² + 25, find the starting height.

For h = −5t(t − 6), when does the object land?

Find the maximum height for h = −5t(t − 6).

More practice: on your own

An object is launched from the ground: h = −4.9t² + 19.6t. Find the maximum height and when it lands.

Water from a fire hose follows h = −0.05d² + 1.2d + 1, where d is the horizontal distance in metres. Find the maximum height of the water and how far away it lands (two decimal places).

Can the ball in h = −5t² + 30t + 1.5 reach 50 m? Explain two ways.

For the diver in Example 2, how high is she after 1 s? Is she above or below the platform?

For the soccer ball in Example 3 (h = −5.12(t − 1.25)² + 8), find its height at t = 0.5 s and the other time it is at that height.

Challenge

A model rocket's height is h = −4.9t² + 49t. For how long is it higher than 100 m?

Ball A is thrown up: h = −5t² + 20t + 1. At the same moment ball B is dropped from a 25 m balcony: h = −5t² + 25. When are they at the same height, and how high?

A ball tossed straight up returns to the thrower's hand after 3 s. Using h = −4.9t(t − 3) for its height above the hand, how high did it go?

Check your understanding

Quiz

For h = −4.9t² + 20t + 1.5, what was the starting height?

For h = −5(t − 3)² + 50, what is the maximum height and when does it occur?

For h = −5t(t − 4), when does the object land?

To find when a ball with height h = −5t² + 20t + 2 is exactly 10 m high, you should solve:

For h = −5t² + 20t + 2, during which time interval is the ball higher than 3 m?

A height model gives roots t ≈ −0.2 and t ≈ 6.1. When does the object land?

For parents and teachers

Big idea: Expectation 4.2 asks students to solve problems arising from a realistic situation represented by a graph or an equation of a quadratic relation, with and without technology. The curriculum's own example is a ball: What is the maximum height? When does it hit the ground? Over what interval is it higher than 3 m? This lesson practises exactly those questions.

How to help: The ball-toss activity connects the math to something your teen can see. Ask them to read each question and say which task it is (vertex, zeros, or solving h = a number) before calculating.

Common difficulties:

Solving h = 0 when the question asks about another height.
Keeping negative times.
Rounding too early in the quadratic formula.
Forgetting that "how long above 3 m" needs the difference between two times.