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 (a) \(t\); (b)...
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
(a) \(t\);
(b) \(t^2\).
Answer Details
For a particle dropped from rest, the initial velocity, \(u=0\). Hence,
\[h=ut+\frac{1}{2}gt^2=\frac{1}{2}gt^2.\]
Taking \(g=9.8\ \text{m s}^{-2}\),
\[h=4.9t^2.\]
(a) Graph of \(h\) against \(t\)
The graph is a parabola passing through the origin. Thus, \(h\) varies parabolically with \(t\), since \(h\propto t^2\). Its gradient increases as time increases.
(b) Graph of \(h\) against \(t^2\)
This is a straight line through the origin. Therefore,
\[h\propto t^2.\]
The gradient of the graph is
\[\frac{h}{t^2}=4.9=\frac{g}{2}.\]
Hence, \(g=2\times4.9=9.8\ \text{m s}^{-2}\).
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