Water Rocket Height Model
A water rocket is launched from a 1.5 m high platform. Its height h metres above the ground after t seconds is modelled by h(t) = −5t² + 20t + 1.5 for 0 ≤ t ≤ 4.5. A sensor records the rocket passing a marker at 18 m on the way up. The launch team wants to know the maximum height, when it occurs, and how long the rocket stays above 18 m, because a camera must be triggered for that interval. A team member claims that because the coefficient of t² is negative, the rocket is highest when h(t) = 0, and another claims it stays above 18 m for over 2 seconds. The teacher asks for a derivative-based check and a quadratic-inequality check before the camera settings are finalised.
1. Find the maximum height of the rocket and the time at which it occurs. Use differentiation and verify by completing the square.
Read the worked answer
Differentiate: h'(t) = −10t + 20. Set h'(t) = 0: −10t + 20 = 0, so t = 2 seconds. This is a maximum because h''(t) = −10 < 0. Then h(2) = −5(2)² + 20(2) + 1.5 = −20 + 40 + 1.5 = 21.5 m. Completing the square: h(t) = −5(t² − 4t) + 1.5 = −5[(t − 2)² − 4] + 1.5 = −5(t − 2)² + 20 + 1.5 = −5(t − 2)² + 21.5. Since −5(t − 2)² ≤ 0, the maximum is 21.5 m at t = 2 s. The claim that the rocket is highest when h(t) = 0 is wrong: h(t) = 0 would mean ground level, not the top of the flight.
2. Find the time interval during which the rocket is above 18 m. Test the claim that it stays above 18 m for over 2 seconds.
Read the worked answer
Solve h(t) > 18: −5t² + 20t + 1.5 > 18. Subtract 18: −5t² + 20t − 16.5 > 0. Multiply by −1, reversing the inequality: 5t² − 20t + 16.5 < 0. Use the quadratic formula on 5t² − 20t + 16.5 = 0: t = [20 ± √(400 − 330)]/10 = [20 ± √70]/10. Since √70 ≈ 8.367, the roots are t ≈ (20 − 8.367)/10 = 1.163 s and t ≈ (20 + 8.367)/10 = 2.837 s. The parabola opens upwards after multiplying by −1, so the inequality holds between the roots: about 1.16 s < t < 2.84 s. The duration is 2.837 − 1.163 = 1.674 s, which is less than 2 seconds. The claim is false; the camera should record for about 1.67 seconds.
3. Find the average rate of change of height from t = 0 to t = 2, and the instantaneous rate of change at t = 1. Compare the two values and interpret.
Read the worked answer
At t = 0, h(0) = 1.5 m. At t = 2, h(2) = 21.5 m. Average rate of change = (21.5 − 1.5)/(2 − 0) = 20/2 = 10 m/s. The instantaneous rate is h'(t) = −10t + 20, so at t = 1, h'(1) = −10 + 20 = 10 m/s. The two values are equal here. This happens because the height is a quadratic with constant acceleration, and the average rate over an interval equals the instantaneous rate at the midpoint of that interval. The value 10 m/s is positive, so the rocket is still rising at t = 1; the vertical speed reaches zero at the maximum height at t = 2.