Model Rocket Height Data
A physics class launches a model rocket straight up and uses a motion sensor to record its height above the ground every second. The data are: at 0 s, 5 ft; at 1 s, 37 ft; at 2 s, 53 ft; at 3 s, 53 ft; at 4 s, 37 ft; at 5 s, 5 ft. One student says the rocket 'must have stopped' between 2 and 3 seconds because the height did not change. Another student proposes the quadratic model h(t) = -8t^2 + 40t + 5, where h is height in feet and t is time in seconds. The class wants to use the model to find the rocket's maximum height and landing time, and to decide whether the model is valid for all t >= 0. The teacher asks them to connect the table, the graph, and the equation rather than just trusting the formula.
1. Verify that h(t) = -8t^2 + 40t + 5 produces the table values, then find the vertex and explain what it means in context. Address the student's 'stopped' claim.
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Substitute the table times: h(0) = 5; h(1) = -8 + 40 + 5 = 37; h(2) = -32 + 80 + 5 = 53; h(3) = -72 + 120 + 5 = 53; h(4) = -128 + 160 + 5 = 37; h(5) = -200 + 200 + 5 = 5. The model matches all six points. For a quadratic in standard form, the vertex time is t = -b/(2a) = -40/(2 * -8) = 2.5 s. The maximum height is h(2.5) = -8(6.25) + 100 + 5 = 55 ft. The rocket reaches 55 ft at 2.5 s. The equal heights at 2 s and 3 s are symmetric about the vertex. The rocket is momentarily at rest at 2.5 s, but it is not stopped for the whole interval from 2 to 3 s; it is moving up before 2.5 s and down after. The average velocity on [2, 3] is zero, but the instantaneous velocity is zero only at the vertex.
2. Solve h(t) = 0 to find when the rocket lands. State the domain for the model and explain why the negative solution is rejected.
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Set -8t^2 + 40t + 5 = 0. Multiply by -1: 8t^2 - 40t - 5 = 0. Using the quadratic formula, t = [40 +/- sqrt(1600 + 160)] / 16 = [40 +/- sqrt(1760)] / 16. Since sqrt(1760) is about 41.95, the two solutions are about (40 + 41.95)/16 = 5.12 and (40 - 41.95)/16 = -0.12. The negative time is rejected because the model describes the rocket after launch, when t >= 0; a negative time would be before the launch and is outside the situation. The rocket lands at about 5.12 s. A reasonable domain is 0 <= t <= 5.12. The model should not be used after landing, because the rocket is on the ground and the quadratic would predict negative heights.
3. Compare the average rates of change on [0, 2.5] and [2.5, 5]. What do the signs and magnitudes tell you about the rocket's motion?
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On [0, 2.5], the average rate of change is [h(2.5) - h(0)] / (2.5 - 0) = (55 - 5)/2.5 = 20 ft/s. On [2.5, 5], it is [h(5) - h(2.5)] / (5 - 2.5) = (5 - 55)/2.5 = -20 ft/s. The positive value means the rocket is gaining height on average during the first interval; the negative value means it is losing height on average during the second. The equal magnitudes reflect the symmetry of the quadratic about the vertex. The average speed over each interval is 20 ft/s, but the instantaneous speed is zero at 2.5 s, when the rocket reverses direction. This pattern is consistent with the table and with the vertex interpretation.