Model Rocket Height, Vertex, and Restricted Inverse
A physics class launches a model rocket from ground level. Its height in feet after t seconds is modeled by h(t) = −16t² + 96t for 0 ≤ t ≤ 6, until it returns to the ground. The class wants to describe the graph, find the maximum height, determine the time interval when the rocket is at least 80 feet high, and decide whether the height function has an inverse on the interval 0 ≤ t ≤ 3. A student says the inverse is h⁻¹(h) = √(h/16) + 3, while another says the domain of the inverse should be all real numbers. The teacher asks for the vertex, intercepts, and a correct restricted inverse with its domain.
1. Find the vertex and intercepts of h(t), and state the domain and range on the given time interval.
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Complete the square: h(t) = −16t² + 96t = −16(t² − 6t) = −16[(t − 3)² − 9] = −16(t − 3)² + 144. The vertex is (3, 144), so the maximum height is 144 ft at t = 3 s. Set h = 0: −16t² + 96t = −16t(t − 6) = 0, so t = 0 and t = 6 are the t-intercepts. The h-intercept is 0. On 0 ≤ t ≤ 6, the domain is [0, 6] and the range is [0, 144].
2. Solve h(t) ≥ 80 and interpret the result in seconds.
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Solve −16(t − 3)² + 144 ≥ 80. Subtract 144: −16(t − 3)² ≥ −64. Divide by −16 and reverse the inequality: (t − 3)² ≤ 4. Thus −2 ≤ t − 3 ≤ 2, so 1 ≤ t ≤ 5. The rocket is at least 80 ft high from t = 1 s to t = 5 s, including both endpoints. This interval is inside the flight time 0 ≤ t ≤ 6.
3. Find the inverse of h on 0 ≤ t ≤ 3, give its domain, and critique the student's formula and domain claim.
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On 0 ≤ t ≤ 3, the graph is one-to-one, so an inverse exists. From h = −16(t − 3)² + 144, rearrange: (t − 3)² = (144 − h)/16, so t = 3 ± sqrt(144 − h)/4. Since t ≤ 3, use the minus sign: h⁻¹(h) = 3 − sqrt(144 − h)/4. The domain of the inverse is the range of h on [0, 3], which is 0 ≤ h ≤ 144. The student's formula √(h/16) + 3 is not correct, and the inverse domain is not all real numbers; h must be restricted to [0, 144].