Reading a parabola from limited data
A ball is launched from a rooftop and its height in metres after t seconds is modelled by h(t) = at^2 + bt + c. Sensors record h(0) = 12, h(2) = 21, and h(5) = 12. A student concludes immediately that the motion is symmetric about t = 2.5 and that the vertex height is therefore h(2.5). A second student objects that three points do not determine a parabola's vertex unless the model is assumed quadratic, and that the symmetry claim needs checking. The launch is known to be upward, so b > 0. The team must decide whether the model is well determined and what the vertex means physically.
1. Determine a, b, c from the three data points and verify or refute the symmetry claim.
Read the worked answer
c = 12 from h(0).
h(2): 4a + 2b + 12 = 21 gives 4a + 2b = 9. h(5): 25a + 5b + 12 = 12 gives 25a + 5b = 0, so 5a + b = 0, b = -5a.
Substitute: 4a + 2(-5a) = 9 gives -6a = 9, a = -1.5, b = 7.5.
So h(t) = -1.5t^2 + 7.5t + 12. Vertex at t = -b/(2a) = 7.5/3 = 2.5, height h(2.5) = -1.5(6.25) + 18.75 + 12 = 21.375.
The symmetry claim is correct: axis t = 2.5, and indeed h(0) = h(5) = 12 are symmetric about 2.5. The vertex height 21.375 m exceeds h(2) = 21, consistent with t = 2 being just before the peak.
2. Three points determine a quadratic only if they are not collinear and no two share the same t. Explain what would go wrong if the third reading had been h(5) = 12 replaced by h(4) = 12, and whether the vertex would still be at 2.5.
Read the worked answer
With h(0) = 12 and h(4) = 12, symmetry about t = 2 would be forced for any quadratic, since equal values at t = 0 and t = 4 place the axis at 2. Using h(2) = 21: 4a + 2b = 9 and 16a + 4b = 0, so 4a + b = 0, b = -4a, giving 4a - 8a = 9, a = -2.25, b = 9, vertex at t = 2 with height 21.
So the vertex moves to 2, not 2.5. The general point: the axis is determined by the pair of equal-height times. Equal values at t = 0 and t = T put the axis at T/2 regardless of the third point, so the vertex location is a consequence of which points coincide in height, not a fixed feature.
3. Using h(t) = -1.5t^2 + 7.5t + 12, find when the ball hits the ground and interpret the negative root.
Read the worked answer
Solve -1.5t^2 + 7.5t + 12 = 0, or 1.5t^2 - 7.5t - 12 = 0, i.e. t^2 - 5t - 8 = 0.
Roots: t = [5 ± sqrt(25 + 32)]/2 = [5 ± sqrt57]/2. sqrt57 is about 7.550, so t ≈ 6.275 or t ≈ -1.275.
The physical impact time is about 6.28 s. The negative root is the time at which the same parabola would have been at ground level before launch; it is a mathematical extension of the model, not an event. Since the model is only claimed for t >= 0, we discard it but note it confirms the algebra: the product of roots is -8 and the sum is 5.