The Balloon Launch Function
A physics class launches a model rocket and records its height over time. The height h(t) in metres after t seconds is modeled by h(t) = −5t² + 40t + 2 for 0 ≤ t ≤ 8. A student claims the rocket reaches its maximum height at t = 8 because 'that is the end of the interval.' Another student graphs the function and says the vertex is at t = 4. The teacher asks the class to determine the maximum height, the time it occurs, and whether the model is reasonable for the entire interval. The class also must decide what the domain restriction means for the range.
1. Find the vertex of h(t) = −5t² + 40t + 2 and interpret it in context. Is the student who says the maximum is at t = 8 correct?
Read the worked answer
For a quadratic in standard form h(t) = at² + bt + c with a = −5, b = 40, c = 2, the vertex occurs at t = −b/(2a) = −40/(2 × −5) = −40/−10 = 4.
h(4) = −5(16) + 40(4) + 2 = −80 + 160 + 2 = 82.
So the vertex is (4, 82): the rocket reaches a maximum height of 82 m at t = 4 s.
Because a < 0, the parabola opens downward, so the vertex is a maximum. The student claiming t = 8 is incorrect: t = 8 is an endpoint, not the vertex. At t = 8, h(8) = −5(64) + 40(8) + 2 = −320 + 320 + 2 = 2 m, which is the landing height, not the maximum.
2. State the range of h(t) on the restricted domain 0 ≤ t ≤ 8, and explain why the range is not simply all values below 82.
Read the worked answer
On 0 ≤ t ≤ 8, evaluate endpoints and vertex:
- h(0) = 2
- h(4) = 82 (maximum)
- h(8) = 2
The function is continuous and the vertex is inside the interval, so the maximum is 82 and the minimum on the interval is 2 (attained at both endpoints).
Range: 2 ≤ h ≤ 82, or [2, 82].
The range is not all values below 82 because the domain is restricted. The function only takes values between the minimum and maximum on that interval. Values above 82 never occur, and values below 2 never occur on 0 ≤ t ≤ 8. The restriction of domain changes the range from the unrestricted parabola's range (h ≤ 82) to the interval [2, 82].
3. The model gives h(8) = 2 m, meaning the rocket is 2 m above the ground at t = 8 s. Does this make the model unreasonable? Explain your reasoning.
Read the worked answer
The model is a mathematical approximation, not a physical law. h(8) = 2 means the model predicts the rocket is 2 m above the launch height at t = 8 s. This could be reasonable if the rocket lands on a platform 2 m high, or if the model is only valid until the rocket hits the ground.
A better interpretation: the model is reasonable only while the rocket is in flight. If the ground is at h = 0, the rocket actually lands when h(t) = 0, which occurs at −5t² + 40t + 2 = 0. Solving: t = [−40 ± √(1600 + 40)]/(−10) = [−40 ± √1640]/(−10). √1640 ≈ 40.5, so t ≈ (−40 − 40.5)/(−10) ≈ 8.05 s. So the model predicts landing just after 8 s, meaning h(8) = 2 is consistent with still being slightly airborne.
The model is reasonable as an approximation, but the domain should ideally end when h = 0. The 2 m offset is small relative to 82 m, so the model is acceptable for the given interval.