Mathematics

Mathematics

Problem

The height h(t)h(t) (in metres) of a stone thrown vertically upwards after tt seconds is given by h(t)=−5t2+20t+3h(t)=-5t^{2}+20t+3 for t≥0t\ge 0. a. (a) Find the value(s) of tt when the stone is at ground level (h=0h=0). b. (b) Determine the maximum height reached by the stone and the time at which it occurs. c. (c) Draw a rough sketch of the graph of hh against tt, marking clearly the intercepts with the axes and the highest point. Explain the physical meaning of the zeros in this context.
Intuitive explanation

The key is recognizing that this is a quadratic function modeling projectile motion, where the −5t2-5t^2 term represents gravity's downward pull and the +20t+20t term captures the initial upward velocity. To find when the stone reaches its maximum height, you don't need to solve an equation—instead, use the fact that a parabola peaks at its vertex, which occurs at t=−b2at = -\frac{b}{2a} for the standard form at2+bt+cat^2 + bt + c. This single calculation unlocks the maximum height and the time it occurs.

Intuitive explanation

The key is recognizing that this is a quadratic function modeling projectile motion, where the −5t2-5t^2 term represents gravity's downward pull and the +20t+20t term captures the initial upward velocity. To find when the stone reaches its maximum height, you don't need to solve an equation—instead, use the fact that a parabola peaks at its vertex, which occurs at t=−b2at = -\frac{b}{2a} for the standard form at2+bt+cat^2 + bt + c. This single calculation unlocks the maximum height and the time it occurs.

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