A particle is dropped from a vertical height h and falls freely for a time t. With the aid of a sketch, explain how h varies with \(t^2\)
Question 1 Report
A particle is dropped from a vertical height h and falls freely for a time t. With the aid of a sketch, explain how h varies with \(t^2\)
Answer Details
Here, \(h\) is most naturally interpreted as the particle’s height above the ground. As the particle falls freely from rest, its downward distance travelled is
\[ s=\tfrac{1}{2}gt^2. \]If its initial height is \(H\), its height remaining above the ground is therefore
\[ h=H-\tfrac{1}{2}gt^2. \]So \(h\) decreases linearly as \(t^2\) increases. A graph of \(h\) against \(t^2\) is a straight line with:
- vertical intercept \(H\), the initial height;
- negative gradient \(-\tfrac{1}{2}g\);
- \(h=0\) when the particle reaches the ground.
Important distinction: if \(h\) were instead defined as the distance fallen, then \(h=\tfrac{1}{2}gt^2\), giving a straight line through the origin with positive gradient \(\tfrac{1}{2}g\). For a question referring to the particle’s vertical height, use \(h=H-\tfrac{1}{2}gt^2\).
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