(i) Zeroes of p(t) = 20t − 16t²
$$p(t) = 20t - 16t^2 = 4t(5 - 4t) = 0$$
$$\Rightarrow t = 0 \quad \text{or} \quad t = \frac{5}{4}$$
The zeroes are 0 and 5/4.
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(ii) Since p(t) = 20t − 16t² is a quadratic polynomial with a negative leading coefficient (−16), its graph is a downward-opening parabola cutting the t-axis at t = 0 and t = 5/4.
→ The correct graph is the one showing a downward parabola with two positive x-intercepts.
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(iii) [Main question]
At $t = \dfrac{3}{2}$:
$$h = 20\left(\frac{3}{2}\right) - 16\left(\frac{3}{2}\right)^2 = 30 - 16 \times \frac{9}{4} = 30 - 36 = -6 \text{ cm}$$
Interpretation: h = −6 cm (negative), which means at t = 3/2 s the dolphin is below the water level. This is not physically possible during the jump, confirming the dolphin re-enters the water before t = 3/2 s (it hits water at t = 5/4 s).
OR
The dolphin hits the water again when h = 0, i.e., at t = 5/4 s (from part i).
The distance covered before hitting water = height function evaluated… The dolphin travels from t = 0 to t = 5/4 s.
Maximum height occurs at $t = \dfrac{5}{8}$ s, $h_{max} = 20\!\cdot\!\frac{5}{8} - 16\!\cdot\!\frac{25}{64} = \frac{25}{4}$ cm.
The total distance covered = 2 × (25/4) = 25/2 = 12.5 cm.
Source: Case Study — Polynomials (Chapter 2)
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