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\(^{2}\).
Question 1 Report
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
Key idea: here \(h\) is the particle’s vertical height above the ground, not the distance it has fallen. If it is released from an initial height \(H\), then
\[h=H-\frac{1}{2}gt^2.\]
(a) Graph of \(h\) against \(t\)
The graph starts at \(h=H\) when \(t=0\). It curves downwards and becomes progressively steeper because the particle’s downward speed increases under the constant acceleration \(g\). It reaches \(h=0\) when the particle reaches the ground.
(b) Graph of \(h\) against \(t^2\)
Rearranging the equation gives
\[h=H+\left(-\frac{g}{2}\right)t^2.\]
This has the form \(y=c+mx\), so the graph is a straight line with:
- vertical intercept \(H\);
- negative gradient \(-\frac{g}{2}\).
Exam reminder: distinguish carefully between height above the ground, \(h=H-\frac12gt^2\), and distance fallen, \(s=\frac12gt^2\). Only distance fallen gives a rising graph through the origin.
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