The Highest Point
Part of the Quadratics & Complex Numbers section of Coddy's Algebra 2 journey. Lesson 42 of 51.
Throw a ball straight up and its height follows a parabola. A ball thrown from 1 metre up has height h(t) = -5t^2 + 20t + 1 metres after t seconds. The -5 comes from gravity, and it is negative, so the parabola opens down. Its vertex is the top of the flight.
Two questions, two steps. When is it highest? The vertex sits at t = -b/(2a), which here is -20/(2(-5)), so 2 seconds. How high is it then? Put that time into the rule: h(2) = -20 + 40 + 1, which is 21 metres.
Keep the two answers apart. The time comes from the formula; the height comes from the rule. A common slip is to report the time as the height.
Challenge
A ball is thrown up from a 2 metre platform. Its height after t seconds is h(t) = -5t^2 + 30t + 2 metres. Find its greatest height h. On the blank board, first write the time of the top as a t = ... line, then the height as an h = ... line.
Test yourselfThis lesson includes a short quiz. Start the lesson to answer it and track your progress.
Test yourselfThis lesson includes a short quiz. Start the lesson to answer it and track your progress.
All lessons in Quadratics & Complex Numbers
2Completing the Square
Perfect Square TrinomialsThe Missing NumberCompleting the SquareIrrational AnswersWhen a Is Not 1Recap: Completing the Square5Complex Arithmetic
Adding and SubtractingMultiplyingThe ConjugateDividingA Pure Imaginary BottomRecap: Complex Arithmetic8Lines Meet Parabolas
Two CrossingsJust TouchingNo MeetingA Circle and a LineRecap: Lines and Parabolas