The Highest Point

Part of the Quadratics & Complex Numbers section of Coddy's Algebra 2 journey. Lesson 39 of 48.

Throw a ball straight up and its height follows a parabola. A ball thrown from 1 metre up has height $h\left(t\right) = -5{t}^{2}+20t+1$ 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 $t = \frac{-b}{2a}$ , which here is $\frac{-20}{2\left(-5\right)}$ , so 2 seconds. How high is it then? Put that time into the rule: $h\left(2\right) = -20+40+1$ , 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 $h\left(t\right) = -5{t}^{2}+30t+2$ 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.