Model Rocket: Quadratic Height and Safety Approval
A physics club launches a model rocket from a 1.5 m high platform on a flat field. The height h metres above the ground after t seconds is modelled by h = 1.5 + 18t − 4.9t² for 0 ≤ t ≤ 4. A drone hovers at a constant height of 12 m on the same vertical line, filming the launch. The club wants to know when the rocket is above the drone, the maximum height reached, and how long the rocket stays above 12 m. They also plan a second launch with a stronger engine, modelled by h = 1.5 + 24t − 4.9t². The safety officer will approve a launch only if the rocket is above the drone for at least 1.8 seconds and the maximum height is below 30 m. The club has a stopwatch that records to the nearest 0.1 s, and they must report times to that precision. They also need to explain why the quadratic graph has the shape it does and what the constant 1.5 represents.
1. For the first model h = 1.5 + 18t − 4.9t², solve h = 12 to find the two times when the rocket is level with the drone. Report times to the nearest 0.1 s and the duration above the drone.
Read the worked answer
Set 1.5 + 18t − 4.9t² = 12. Rearrange: 4.9t² − 18t + 10.5 = 0. Discriminant = 18² − 4(4.9)(10.5) = 324 − 205.8 = 118.2. √118.2 = 10.872. t = (18 ± 10.872) / 9.8. So t = 0.727 s or t = 2.946 s. To nearest 0.1 s: 0.7 s and 2.9 s. Duration = 2.946 − 0.727 = 2.219 s ≈ 2.2 s. The rocket is above 12 m between these times because the quadratic opens downwards (coefficient −4.9 < 0). The constant 1.5 is the launch height above ground, not part of the motion generated by the engine.
2. Find the maximum height reached by the first model and the time at which it occurs. State whether the safety officer's 30 m limit is met.
Read the worked answer
For h = −4.9t² + 18t + 1.5, maximum occurs at t = −b/(2a) = −18/(2 × −4.9) = 18/9.8 = 1.837 s. Substitute: h = 1.5 + 18(1.837) − 4.9(1.837)². 18 × 1.837 = 33.066; 1.837² = 3.375; 4.9 × 3.375 = 16.538; h = 1.5 + 33.066 − 16.538 = 18.028 m, about 18.0 m. This is below 30 m. Since the rocket is above 12 m for 2.2 s, which is at least 1.8 s, the first launch meets both safety conditions.
3. Repeat the key checks for the second model h = 1.5 + 24t − 4.9t². Would the safety officer approve it? Support your answer with calculations.
Read the worked answer
Solve 1.5 + 24t − 4.9t² = 12: 4.9t² − 24t + 10.5 = 0. Discriminant = 576 − 205.8 = 370.2; √370.2 = 19.24. t = (24 ± 19.24)/9.8 = 0.486 s or 4.412 s. Duration ≈ 3.9 s, which passes the 1.8 s condition. Maximum at t = 24/9.8 = 2.449 s. h = 1.5 + 24(2.449) − 4.9(2.449)² = 1.5 + 58.776 − 29.389 = 30.887 m, about 30.9 m. This exceeds the 30 m limit, so the safety officer should not approve the second launch even though it stays above the drone longer. The limiting condition is maximum height.