Lesson 37: Applications of Quadratics (Projectiles, Area Problems)
Quadratic functions appear in real-life situations such as projectile motion and area problems. Understanding how to model these scenarios with quadratics allows us to solve practical problems.
Projectile Motion
The height of an object in projectile motion can be modeled as:
h(t) = −16t² + vt + s
- h(t) = height at time t (in feet)
- v = initial velocity (ft/s)
- s = initial height (ft)
- −16 comes from gravity acceleration in ft/s²
Example:
An object is thrown upward with an initial velocity of 32 ft/s from a height of 48 ft. Find maximum height and time to hit the ground.
- h(t) = −16t² + 32t + 48
- Vertex: t = −b/(2a) = −32/(2 * −16) = 1 second → maximum height h(1) = −16(1)² + 32(1) + 48 = 64 ft
- Set h(t) = 0 to find when it hits ground: −16t² + 32t + 48 = 0 → divide by −16: t² − 2t − 3 = 0 → (t − 3)(t + 1) = 0 → t = 3 seconds (positive solution)
Area Problems
Quadratics can model the area of rectangles where one side depends on another.
Example:
A rectangle has a perimeter of 40 meters. Express area as a quadratic in terms of length x.
- Perimeter formula: 2x + 2y = 40 → y = 20 − x
- Area A = x * y = x(20 − x) = −x² + 20x (standard form)
- Vertex: x = −b/(2a) = −20/(2 * −1) = 10 → maximum area A = −(10)² + 20*10 = 100 m²
Practice Problems
- An object is thrown upward with velocity 48 ft/s from 64 ft. Find maximum height and time to hit ground.
- A rectangle has perimeter 50 m. Find dimensions for maximum area.
- Height of a ball is given by h(t) = −16t² + 24t + 8. When does it reach maximum height and what is that height?
- Find maximum area for a rectangle with perimeter 60 m.
- Projectile launched at 40 ft/s from 32 ft. Determine height at t = 1s and when it hits the ground.