O-Level Quadratic functions: worked solution

Question

A ball thrown vertically upward has height h=7+20t−5t² m at time t s. (a) Find the greatest height reached and when it happens.

h = 7 + 20t - 5t² is a quadratic in t. The coefficient of t² is -5, which is negative, so the graph is an upside-down parabola and the ball's height has a MAXIMUM at the vertex. Completing the square finds that vertex. First put the terms in t at the front and take out the factor -5 from them (leaving the constant 7 outside): h = -5t² + 20t + 7 h = -5(t² - 4t) + 7. Check that step by expanding: -5 × t² = -5t² and -5 × (-4t) = +20t, correct. Now complete the square inside the bracket. Halve the coefficient of t: half of -4 is -2, so the bracket to use is (t - 2)². Expanding, (t - 2)² = t² - 4t + 4, which is 4 too big, so t² - 4t = (t - 2)² - 4. Substitute that in: h = -5[(t - 2)² - 4] + 7 h = -5(t - 2)² + 20 + 7 h = -5(t - 2)² + 27. Now read off the maximum. A square is never negative, so (t - 2)² ≥ 0, which means -5(t - 2)² ≤ 0. Therefore h is at its largest when -5(t - 2)² is zero, i.e. when t - 2 = 0, so t = 2. At t = 2: h = -5(0) + 27 = 27. Greatest height = 27 m, reached at t = 2 s.