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)...
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) \(t^2\).
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 \(t\)
Graph of height fallen, h, against time, t. The curve is parabolic.
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 \(t^2\)
Graph of height fallen, h, against t². The straight line passes through the origin and has gradient 4.9 m s⁻².
This is a straight line through the origin. Therefore,
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 \(t\)
Graph of height fallen, h, against time, t. The curve is parabolic.
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 \(t^2\)
Graph of height fallen, h, against t². The straight line passes through the origin and has gradient 4.9 m s⁻².
This is a straight line through the origin. Therefore,