Figure9.1.A particle projected from \(O\) with speed \(v_0\) at elevation \(\theta\text{.}\) At the point \(P\) reached after time \(t\text{,}\) the velocity \(v\) is tangent to the path and makes an angle \(\phi\) with the horizontal; \(h\) is the depth of \(P\) below the directrix.
A particle is projected from \(O\) with speed \(v_0\) at an elevation \(\theta\) to the horizontal, as in Figure 9.1. At time \(t\) it is at the point \(P\text{,}\) moving with speed \(v\) in a direction making an angle \(\phi\) with the horizontal. Show that
\begin{equation*}
v = \sqrt{\,v_0^2 - 2\,v_0\,g\,t\sin\theta + g^2 t^2\,}
\end{equation*}
The directrix of the parabolic path is the horizontal line at height \(\dfrac{v_0^2}{2g}\) above \(O\text{.}\) Using the previous exercise, show that the speed of the particle at \(P\) is equal to the speed it would acquire by falling freely from the level of the directrix down to \(P\text{.}\) That is, if \(h\) denotes the depth of \(P\) below the directrix (see Figure 9.1), then \(v^2 = 2\,g\,h\text{.}\)
