A particle dropped from a vertical height and falls freely for a time interval t. Sketch and explain a graph to show how h varies with (a) t (b) t\(^{2}\).

Answer Details

Take h as the height of the particle above the ground and let it be released from rest from a height H. Since the fall is free (only gravity acts), the distance fallen after time t is \(\tfrac{1}{2}gt^{2}\), so the height still remaining above the ground is

\[ h = H - \tfrac{1}{2}\,g\,t^{2}. \]

Using \(H = 45\ \text{m}\) and \(g = 10\ \text{m s}^{-2}\) as a worked illustration, \(h = 45 - 5t^{2}\). The values are:

(a) How h varies with t

Because \(h = H - \tfrac{1}{2}gt^{2}\), the term subtracted grows as the square of the time. The graph therefore starts at \(h = H\) on the vertical axis and curves downward (a parabola, concave down). Its slope, which is the negative of the speed \(\left(\dfrac{dh}{dt} = -gt\right)\), is zero at the instant of release and becomes steadily steeper as the particle accelerates, reaching the ground (\(h = 0\)) after \(t = 3\ \text{s}\). So h falls slowly at first and ever faster afterwards, giving a curved (non-linear) graph.

(b) How h varies with t\(^{2}\)

Writing \(x = t^{2}\), the relation becomes \(h = H - \tfrac{1}{2}g\,x\), which is of the linear form \(y = c + mx\). The graph of h against \(t^{2}\) is therefore a straight line with

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• intercept on the h-axis equal to the release height, \(c = H = 45\ \text{m}\);
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• a negative slope equal to \(-\tfrac{1}{2}g\).
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Reading the slope from the line, \[ \text{slope} = \frac{0 - 45}{9 - 0} = -5\ \text{m s}^{-2} = -\tfrac{1}{2}g, \] which gives \(g = 10\ \text{m s}^{-2}\). Thus plotting h against \(t^{2}\) straightens the curve of part (a) into a line whose gradient conveniently yields the acceleration due to gravity.