Lesson 18: Quadratic Applications: Projectile Motion and Optimization

Quadratics model anything that rises then falls — or falls then rises — symmetrically: thrown balls, arched bridges, revenue that peaks at a sweet-spot price. This lesson is the application workshop for Lessons 10–11: vertex for the extreme value, roots for the start and finish, and optimisation for the best possible plan.

Projectiles: the −16t² family

With height in feet and time in seconds, gravity gives h(t) = −16t² + v₀t + h₀, where v₀ is the initial upward velocity and h₀ the launch height. Each constant has a job: h₀ is where the story starts, v₀ how hard it was thrown, −16 the pull of gravity. Worked example 1. h(t) = −16t² + 64t (thrown from the ground). Factor: −16t(t − 4), so the roots are t = 0 (launch) and t = 4 s (landing). The vertex is midway between the roots — a handy consequence of symmetry — at t = 2, and h(2) = −64 + 128 = 64 ft maximum. The height table shows the symmetry physically:

Worked example 2. h(t) = −16t² + 32t + 80: a ball thrown upward at 32 ft/s from the top of an 80 ft platform. Maximum at t = −32/(−32) = 1 s: h(1) = −16 + 32 + 80 = 96 ft. Landing: −16t² + 32t + 80 = 0; divide by −16: t² − 2t − 5 = 0; the formula gives t = (2 ± √24)/2 = 1 ± √6. The meaningful root: 1 + √6 ≈ 3.45 s. The other, 1 − √6 ≈ −1.45 s, is a time before the throw — the mathematics offers it, the context forbids it. This is the same reject-the-impossible-root discipline as factoring problems in Lesson 9.

Optimisation: the perimeter family

Worked example 3. A farmer has 40 m of fencing for a rectangular pen. What dimensions maximise the area? With width x, the length is (40 − 2x)/2 = 20 − x, so A(x) = x(20 − x) = −x² + 20x — a downward parabola. Vertex at x = 20/2 = 10 m, length also 10 m: the square, with area 100 m². Neighbours confirm: 8 × 12 gives 96, 12 × 8 gives 96, 5 × 15 gives 75. The moral generalises: for fixed perimeter, the square is the best rectangle — the vertex of the area parabola always lands on it. Revenue optimisation runs identically. A stall sells 400 candles at $20; for each $1 price rise it loses 10 sales. With x = number of $1 rises, price is 20 + x and sales 400 − 10x, so R(x) = (20 + x)(400 − 10x) = −10x² + 200x + 8000. Vertex at x = −200/(−20) = 10 → price $30, sales 300, revenue R = 30 × 300 = $9,000 (formula check: −1000 + 2000 + 8000 ✓). At x = 0 and x = 20 the revenue is $8,000 both times — symmetric around the vertex, as it must be.

Choosing the tool for the question asked

• "Maximum height / best price / largest area" → vertex (x = −b/(2a), then evaluate).
• "When does it land / break even / hit zero" → roots (factoring or the formula; reject impossible roots in context).
• "Height at time t" → plain evaluation.

Misreading which question is asked is the most common way to lose an otherwise-perfect solution: the vertex's x-coordinate answers "when/what price", its y-coordinate answers "how high/how much" — and problems regularly ask for one while students proudly deliver the other.

Modelling application: one scenario, three questions

For h(t) = −16t² + 48t + 16: the maximum height is at t = 1.5 s and equals h(1.5) = −36 + 72 + 16 = 52 ft; the ball lands when −16t² + 48t + 16 = 0, i.e. t = (48 + √(2304 + 1024))/32 = (48 + √3328)/32 ≈ (48 + 57.7)/32 ≈ 3.30 s; and the launch height — the answer to "where did it start?" — is simply h(0) = 16 ft, the constant. Three questions, three different features of the same parabola. Drawing the curve and labelling vertex, intercepts and launch point before computing anything is the professional habit that keeps the answers sorted.

Common mistakes

• Delivering the wrong vertex coordinate. If the question is "what is the maximum height?", the answer is 52 ft (the y-value), not 1.5 s (the time). State both in your working, then choose.
• Accepting the negative time root. 1 − √6 seconds happened *before* the throw. Negative lengths, negative counts and past times are the model's way of offering algebra without physics — decline.
• Using the full perimeter when a wall forms one side. If a pen is built against a barn, the fence covers only three sides: 2 widths + 1 length = total fence. Drawing the picture — with the wall drawn in — prevents the error; the Exam Experience below tests exactly this.
• Rounding mid-solution. Keep √6 symbolic until the final step: (48 + √3328)/32 with √3328 ≈ 57.69 gives 3.30 s; premature rounding to √3328 ≈ 58 gives 3.31 — small, but chained approximations grow. Evaluate exactly, round once, at the end.

Exam Experience

Question 1. The height of a ball is h(t) = −16t² + 48t + 16 feet. Its maximum height is: A. 1.5 ft B. 52 ft C. 16 ft D. 48 ft Answer: B. The vertex is at t = 48/32 = 1.5 s (option A is the *time*), and h(1.5) = −36 + 72 + 16 = 52 ft. The question asks for height; 16 ft (C) is the launch height.

Question 2. A ball with height h(t) = −16t² + 64t returns to the ground at: A. t = 2 s B. t = 8 s C. t = 4 s D. t = 64 s Answer: C. −16t(t − 4) = 0 gives roots 0 and 4. (t = 2, option A, is the *peak* time — the vertex sits midway between the roots by symmetry.)

Question 3. A rectangle has perimeter 24 m. The greatest possible area is: A. 24 m² B. 36 m² C. 35 m² D. 144 m² Answer: B. A = x(12 − x), vertex at x = 6: the 6 × 6 square, area 36 m². Option C (5 × 7) is close but beaten by the square; D confuses squaring the perimeter with maximising area.

Question 4 (constructed response). A gardener has 60 ft of fencing to enclose a rectangular vegetable plot against an existing wall, so fencing is needed on only three sides. Let x be the length (in feet) of each side perpendicular to the wall. (a) Write the area A(x) as a function of x. (b) Find the dimensions that maximise the area, and the maximum area. (c) The gardener wants an area of exactly 400 ft². Find all possible values of x, and give the corresponding plot dimensions for each.

Scoring guide (4 points).

• 1 point: the side parallel to the wall is 60 − 2x, so A(x) = x(60 − 2x) = −2x² + 60x.
• 1 point: vertex at x = 60/4 = 15 ft, parallel side 60 − 30 = 30 ft, maximum area 15 × 30 = 450 ft².
• 2 points: solving x(60 − 2x) = 400 → x² − 30x + 200 = 0 → (x − 10)(x − 20) = 0 gives x = 10 or x = 20 (1 point); with dimensions 10 ft × 40 ft and 20 ft × 20 ft respectively (1 point) — both valid, a useful reminder that an area equation can have two genuinely different gardens as answers.

Deeper still: "does it ever reach…?" — the discriminant as a yes/no oracle

Some application questions ask not for a maximum but for *possibility*. The ball of Lesson 10, h(t) = −16t² + 32t + 6, peaked at 22 ft. Does it ever reach 25 ft? Set the height equal to 25: −16t² + 32t + 6 = 25 → −16t² + 32t − 19 = 0. Discriminant: 32² − 4(−16)(−19) = 1024 − 1216 = −192 < 0 — no real solutions, so the ball *never* reaches 25 ft, agreeing with the maximum of 22 ft. Two methods (vertex height vs discriminant sign), one fact: they are the same question viewed from the graph and from the equation. The pattern is worth naming: "does the output ever equal K?" always becomes "solve f(t) = K and interrogate the discriminant" — K below the maximum of a downward parabola gives two times (once rising, once falling), K equal to it gives one, K above gives none. Break-even analysis in Lesson 8 is the same oracle in business dress: a price at which revenue equals cost exists only if the relevant quadratic actually crosses zero. Change the planet and the family persists: on the Moon, gravity's coefficient is about −2.7 instead of −16, so the same throw follows h(t) = −2.7t² + 32t + 6: the peak (t ≈ 5.9 s, height ≈ 101 ft) and the flight time (≈ 12.0 s) balloon — one changed constant, an entirely different afternoon. Constants are the model's fingerprints of place and context; the algebra doesn't blink.

One more idea: parabolas in architecture — the bridge arch

A bridge's arch is often a parabola. Suppose an arch follows y = −x²/20 + 5, with x measured horizontally from the centre and y the height above the road's baseline (metres). The vertex (0, 5) is the arch's crown, 5 m up. The arch meets the baseline where −x²/20 + 5 = 0 → x² = 100 → x = ±10: a span of 20 m. A truck passing 5 m off-centre has clearance y = −25/20 + 5 = 3.75 m — which is why tall vehicles are routed to the centre lane: on a parabolic arch, clearance is a quadratic function of lane position, falling slowly near the crown and faster near the edges. Engineers choose the coefficient (−1/20 here) to trade span against height; the algebra you have been practising — vertex for the peak, roots for the span, evaluation for clearance at a point — is literally the design calculation, and the same three questions from the projectile problems (how high, how far, what value here) organise it completely.