Quadratic word problems: height and area

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Learn it For a thrown object, h(t) = −4.9t² + vt + h₀ (metres, seconds): the vertex gives the maximum height and its time, and the positive root gives when it lands. For area problems, write the area as a product, set it equal to the target, and solve.

Example: h(t) = −5t² + 20t + 1. Find the maximum height.

t at the vertex = −b ÷ 2a = −20 ÷ (−10) = 2 s.
h(2) = −20 + 40 + 1 = 21 m.

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