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Mathematics 🌆 Grade 9 Projectile Launcher: Roots and Vertex of a Quadratic Model
🌆 Grade 9 · Lesson 6 of 12

Projectile Launcher: Roots and Vertex of a Quadratic Model

A ball tossed into the air traces a parabola — its roots are the moments it touches the ground, and its vertex is the peak of the flight.

Grade 9Algebra 1
Projectile Launcher: Roots and Vertex of a Quadratic Model — illustration
💡
The big idea: When something is launched into the air, its height over time follows a quadratic function. The function's roots (where height equals zero) mark launch and landing, and its vertex marks the highest point of the flight. Factoring and the vertex formula turn a physical flight into numbers you can compute exactly.
🎯 By the end, you'll be able to
  • Model the height of a launched object with a quadratic function of time
  • Find when an object lands by finding the roots of the height function
  • Find the time and height of the peak using the vertex formula t = −b ÷ (2a)
  • Interpret the roots and vertex of a projectile model in real-world terms
📎 You should already know
  • Factoring quadratics
  • Evaluating algebraic expressions

A flight traced in numbers

Throw a ball straight up, launch a firework, or kick a football, and its height over time traces a smooth arc — a parabola. That arc can be described exactly by a quadratic function h(t), where t is the time since launch and h(t) is the height at that moment.

🔑 Roots are ground level, vertex is the peak
The roots of h(t) are the times when the height is 0 — the object is at ground level (launch and landing). The vertex of h(t) is the highest point of the arc — the moment and height of the peak.
\[ h(t) = at^2 + bt + c \]
A quadratic model of height over time. For an object launched from the ground with only gravity acting on it, a is negative (the arc opens downward).
🎮 Projectile Launcher LIVE
Set launch speed and angle; the parabola's roots are the landing points and the vertex is the peak height.

Finding the vertex: time and height of the peak

For any quadratic at² + bt + c, the vertex occurs at t = −b ÷ (2a). Once you know that time, plug it back into h(t) to find the peak height itself.

\[ t_{\text{peak}} = \dfrac{-b}{2a} \]
The time of the peak height, straight from the coefficients of the quadratic.

Finding the roots: where the object lands

Setting h(t) = 0 and factoring (or using the quadratic formula) gives the times when the height is zero. One root is usually t = 0 (the launch); the other positive root is the landing time.

📝 Worked example: A ball is launched from the ground with h(t) = −16t² + 64t (t in seconds, h in feet). When does it land?
  1. Set the height to 0: −16t² + 64t = 0.
  2. Factor out −16t: −16t(t − 4) = 0.
  3. So t = 0 (the launch) or t = 4 (the landing).
✓ The ball lands at <strong>t = 4 seconds</strong>.
📝 Worked example: For the same ball, h(t) = &minus;16t&sup2; + 64t, find the time and height of its peak.
  1. Use t = −b ÷ (2a) with a = −16, b = 64: t = −64 ÷ (2 × −16) = −64 ÷ −32.
  2. So the peak occurs at t = 2 seconds.
  3. Substitute t = 2 into h(t): h(2) = −16(2)² + 64(2) = −16(4) + 128 = −64 + 128.
✓ The peak height is <strong>64 feet</strong>, reached at <strong>t = 2 seconds</strong>.
✨ The vertex sits exactly between the roots
Notice the ball landed at t = 4, and its peak came at t = 2 — exactly halfway between the roots t = 0 and t = 4. A parabola is symmetric, so its vertex always falls midway between any two points where it has equal height, including its roots.
⚠️ Watch the sign of a
For a projectile launched upward under gravity, a is negative (the parabola opens downward, so it has a maximum, not a minimum). Dropping the negative sign when computing t = −b ÷ (2a) is a common source of errors — double check the sign before you divide.

Check your understanding

1. For h(t) = &minus;16t&sup2; + 48t, at what positive time does the object land?
Factor: −16t(t − 3) = 0, so t = 0 or t = 3. The object lands at t = 3 seconds.
2. For h(t) = &minus;16t&sup2; + 48t, at what time does the peak height occur?
t = −b ÷ (2a) = −48 ÷ (2 × −16) = −48 ÷ −32 = 1.5 seconds.
3. What is the peak height for h(t) = &minus;16t&sup2; + 48t?
h(1.5) = −16(1.5)² + 48(1.5) = −16(2.25) + 72 = −36 + 72 = 36 feet.
4. In the vertex formula t = −b ÷ (2a), the letters a and b come from which form of the quadratic?
The formula t = −b ÷ (2a) uses the coefficients a and b from standard form, ax² + bx + c.
5. A height function has two distinct positive roots, t₁ and t₂. What do these roots represent?
Roots of a height function are the times when h(t) = 0 — ground level, such as the moments of launch and landing.
✅ Key takeaways
  • A launched object's height over time follows a quadratic function h(t).
  • The roots of h(t) are when the height is zero — typically the launch and landing times.
  • The vertex of h(t) gives the time and value of the peak height, found with t = −b ÷ (2a).
  • The peak time always falls exactly halfway between two times of equal height, including the roots.
  • For an upward launch under gravity, the leading coefficient a is negative, since the arc opens downward.