Mathematics

Mathematics

Problem

A stone is projected vertically upward from the ground. Its height hh metres after tt seconds is given by the expression h=20t−5t2h=20t-5t^{2}. (i) After how many seconds will the stone return to the ground? (Take the instant of projection as t=0t=0.) (ii) At what time will the stone reach its maximum height? (iii) What is this maximum height?
Intuitive explanation

The key is recognizing that h=20t−5t2h = 20t - 5t^2 is a quadratic function describing the stone's position over time. For part (i), the stone returns to ground level when h=0h = 0—this gives you a factorizable equation with two solutions (one at launch, one at landing). For parts (ii) and (iii), the maximum occurs at the vertex of the parabola, which you can find using calculus (set dhdt=0\frac{dh}{dt} = 0) or the symmetry property that the peak happens halfway through the flight.

Intuitive explanation

The key is recognizing that h=20t−5t2h = 20t - 5t^2 is a quadratic function describing the stone's position over time. For part (i), the stone returns to ground level when h=0h = 0—this gives you a factorizable equation with two solutions (one at launch, one at landing). For parts (ii) and (iii), the maximum occurs at the vertex of the parabola, which you can find using calculus (set dhdt=0\frac{dh}{dt} = 0) or the symmetry property that the peak happens halfway through the flight.

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