The Rocket's Two Roots
A model rocket is launched vertically from a platform. Its height in metres after t seconds is modelled by h(t) = −5t² + 30t + 35 for 0 ≤ t ≤ 8. A sensor records the height at several times, and a student plots the graph, noticing it is a downward parabola with vertex somewhere between t = 2 and t = 4. The class is asked to extract maximum height, the time of flight until the rocket returns to ground level, and the interval during which the rocket is above 60 m. A second student claims the rocket is above 60 m for exactly 2 seconds and wants this checked.
1. Find the maximum height and the time at which it occurs, and find when the rocket hits the ground.
Read the worked answer
Complete the square: h(t) = −5(t² − 6t) + 35 = −5[(t − 3)² − 9] + 35 = −5(t − 3)² + 45 + 35 = −5(t − 3)² + 80.
Vertex at t = 3, maximum height 80 m. This lies in [0, 8], so it is attained.
Ground level: −5t² + 30t + 35 = 0, divide by −5: t² − 6t − 7 = 0, so (t − 7)(t + 1) = 0, giving t = 7 (reject t = −1). The rocket lands at t = 7 s.
2. Solve h(t) > 60 and evaluate the second student's claim.
Read the worked answer
−5t² + 30t + 35 > 60 ⇒ −5t² + 30t − 25 > 0 ⇒ divide by −5 and flip: t² − 6t + 5 < 0 ⇒ (t − 1)(t − 5) < 0 ⇒ 1 < t < 5.
The rocket is above 60 m for 5 − 1 = 4 seconds, not 2. The student's claim is wrong; the interval is (1, 5), of length 4 s. The symmetry about t = 3 confirms this: the two crossings at t = 1 and t = 5 are each 2 s from the vertex time.