A. Quadratic Functions · Quadratic Functions

Solving Problems: Heights, Motion and Arches

A3.3 A2.4About 70 minutes

Learning goals

I can translate questions about heights and arches into equations.
I can find a maximum height and when it occurs.
I can find when an object reaches a given height or lands, and for how long it is above a height.
I can answer in context with sensible rounding and units.

Words to know

projectile
An object thrown or launched that moves under gravity alone (ignoring air resistance).

maximum height
The k-value of the vertex of a height function that opens down.

time in the air
The positive zero of the height function (from launch at t = 0).

quadratic formula
x = (−b ± √(b² − 4ac)) ÷ (2a), the solutions of ax² + bx + c = 0.

Questions a model can answer

Once you have a quadratic function for a moving object, you can answer real questions with it: How high does it go? When? When does it land? For how long is it above the net? Will it clear the tree? These are exactly the questions engineers, coaches and game designers ask.

In this lesson the model is given. Your job is to translate each question into mathematics, choose the right tool, and answer in context.

Question → tool

Maximum height and when it happens → the vertex: complete the square, use h = −b ÷ (2a), or use the zeros.
Height at a given time → evaluate h(t).
Starting height → h(0), the constant term.
When it lands → solve h(t) = 0. Factor if possible; otherwise use the quadratic formula from Quadratic Expressions and Equations. Reject negative times.
When it is at a given height k → solve h(t) = k. There are usually two answers: once going up, once coming down.
For how long it is above height k → the difference between those two times.

Always check the domain (t ≥ 0 and before landing), round sensibly, and write a sentence with units.

Remember the physics

For an object moving freely near Earth's surface (ignoring air resistance), height in metres after t seconds is h(t) = −4.9t² + v₀t + h₀, where v₀ is the starting upward speed in m/s and h₀ is the starting height in metres. Some textbooks round −4.9 to −5 to make the numbers easier.

Worked example

Example 1: A volleyball serve

A volleyball is served upward. Its height is h(t) = −4.9t² + 9.8t + 1.5 metres after t seconds.

(a) Find the maximum height and when it occurs. (b) For how long is the ball higher than 5 m? (c) If nobody touches it, when does it hit the floor?

(a) h = −b ÷ (2a) = −9.8 ÷ (−9.8) = 1 s. h(1) = −4.9 + 9.8 + 1.5 = 6.4 m.
(b) Solve −4.9t² + 9.8t + 1.5 = 5, which is −4.9t² + 9.8t − 3.5 = 0.
t = (−9.8 ± √(9.8² − 4(−4.9)(−3.5))) ÷ (2(−4.9)) = (−9.8 ± √(96.04 − 68.6)) ÷ (−9.8) = (−9.8 ± √27.44) ÷ (−9.8).
√27.44 ≈ 5.238, so t ≈ 0.465 or t ≈ 1.535. The ball is above 5 m for about 1.535 − 0.465 ≈ 1.07 s.
(c) Solve −4.9t² + 9.8t + 1.5 = 0: t = (−9.8 ± √(96.04 + 29.4)) ÷ (−9.8) = (−9.8 ± √125.44) ÷ (−9.8) = (−9.8 ± 11.2) ÷ (−9.8).
t ≈ −0.14 (reject: before the serve) or t = 21 ÷ 9.8 ≈ 2.14 s.

Maximum 6.4 m at 1 s; above 5 m for about 1.07 s; hits the floor at about 2.14 s.

The serve h(t) = −4.9t² + 9.8t + 1.5. The horizontal line h = 5 crosses the curve at about 0.47 s and 1.53 s; the shaded band shows the time the ball is above 5 m.

Worked example

Example 2: A platform dive

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

Vertex: t = −3 ÷ (2(−4.9)) = 3 ÷ 9.8 ≈ 0.31 s.
h(0.306) ≈ −4.9(0.0937) + 0.918 + 10 ≈ 10.46 m, about half a metre above the platform.
Water: −4.9t² + 3t + 10 = 0. t = (−3 ± √(9 + 196)) ÷ (−9.8) = (−3 ± √205) ÷ (−9.8) = (−3 ± 14.318) ÷ (−9.8).
t ≈ −1.15 (reject) or t ≈ 17.318 ÷ 9.8 ≈ 1.77 s.

Maximum about 10.46 m at about 0.31 s; she enters the water at about 1.77 s.

Worked example

Example 3: Will it fit under the arch?

A bridge has a parabolic arch modelled by f(x) = −0.025(x − 40)² + 40, where x is the horizontal distance in metres from one end at water level, and f(x) is the height in metres. A barge carrying a load 30 m tall must pass under the arch. How wide can the load be?

The arch is at least 30 m high where f(x) ≥ 30.
Solve f(x) = 30: −0.025(x − 40)² + 40 = 30, so (x − 40)² = 10 ÷ 0.025 = 400.
x − 40 = ±20, so x = 20 or x = 60.
Between x = 20 and x = 60 the arch is higher than 30 m. That is a width of 40 m, centred under the top of the arch.
(The arch meets the water where (x − 40)² = 1600: at x = 0 and x = 80, so it spans 80 m.)

A 30 m tall load can be at most 40 m wide, if it is centred.

The arch f(x) = −0.025(x − 40)² + 40 spans 80 m and is 40 m high in the middle. It is at least 30 m high only between x = 20 and x = 60.

Just for fun

Ignoring air resistance, a ball thrown straight up comes back to your hand at the same speed it left with. It just changed its mind about the direction.

Hands-on activity

Measure a throw with a stopwatch

You need: a soft ball (a tennis ball or a beanbag), a phone stopwatch or video, a partner, and an open space outdoors or in a gym, away from windows, people and anything breakable.

Stand still and throw the ball straight up. Your partner times how long it takes to come back to the same height (your hand). Call this time T. Do 3 throws and average T.
If the ball leaves and returns at the same height, its height above your hand is h(t) = −4.9t(t − T). Explain why the zeros are t = 0 and t = T.
The top is halfway: t = T ÷ 2. Show that the maximum height above your hand is 4.9 × (T ÷ 2)².
Compute the maximum height of your best throw. Then the starting speed: v₀ = 4.9T (in m/s).
Optional: record a video with the ball next to a tall wall or building with a known height (or a metre stick for low throws) and compare.

What to look for: a throw with T = 2.0 s gives a maximum height of 4.9 × 1² = 4.9 m above the hand and a starting speed of 9.8 m/s. Measured times have reaction-time error of about 0.1 to 0.2 s, so the heights are estimates.

Your turn

A ball's height is h(t) = −5t² + 30t + 2 metres. Find the maximum height.

For the same ball, find the height at t = 5 s.

When does that ball hit the ground? Round to two decimal places.

h(t) = −4.9t² + 19.6t. When is the object 14.7 m high?

h(t) = −5(t − 2)² + 22. What is the starting height, and when does the ball land?

Will a ball with h(t) = −4.9t² + 10t + 1 reach a height of 7 m? Explain.

More practice: warm-up

h(t) = −5t² + 20t. Find h(1).

State the maximum height and when it occurs for h(t) = −5(t − 3)² + 50.

To find when an object lands, which equation do you solve?

What is the starting height for h(t) = −4.9t² + 12t + 1.8?

Solving h(t) = 0 gives t = −0.4 and t = 3.2. Which answer do you keep?

More practice: on your own

A springboard diver's height is h(t) = −4.9t² + 4.9t + 3 metres. Find the maximum height and when the diver enters the water.

For h(t) = −5t² + 30t + 2, for how long is the ball above 20 m?

A tunnel entrance is modelled by h(x) = −0.5x² + 4x (metres). How wide and how high is it? Can a truck 3 m wide and 6 m tall drive through the middle?

Ball A: h = −5t² + 20t + 1. Ball B: h = −5t² + 15t + 10. Which goes higher? When are they at the same height?

A rocket's height is h(t) = −4.9t² + 49t + 1 metres. Find its maximum height and how long it stays above 100 m.

Challenge

A ball thrown from 1.5 m reaches a maximum of 21.5 m after 2 s. Write its height function in vertex form and find when it lands.

An object is launched from the ground with h(t) = −4.9t² + vt. Show that it is in the air for v ÷ 4.9 seconds and reaches a maximum height of v² ÷ 19.6 metres.

For the bridge f(x) = −0.025(x − 40)² + 40, a load is 20 m wide and centred. How tall can it be?

Check your understanding

Quiz

h(t) = −5t² + 40t. What is the maximum height?

For h(t) = −5t² + 40t, when does the object land?

What is the starting height for h(t) = −4.9t² + 15t + 2?

For h(t) = −5t² + 20t, when is the height 15 m?

Why do we usually reject negative roots of h(t) = 0?

What are the maximum height and its time for h(t) = −5(t − 1.5)² + 12?

For parents and teachers

Big idea: expectation A3.3 asks students to solve problems arising from real-world applications, given the algebraic representation of a quadratic function. This lesson focuses on heights, motion and arches.

How to help: for each problem, ask your teen to write the question as an equation first ("when is it 5 m high?" becomes h(t) = 5) before calculating. Insist on a final sentence with units.

Common difficulties:

Solving h(t) = 0 when the question asks for a different height.
Giving only one of two times for a given height.
Arithmetic errors in the quadratic formula with negative a: write every step.
Rounding too early: keep extra decimals until the final answer.
Keeping negative times.