1.
A particle is projected from \(O\) with speed \(v_0\) at an elevation \(\theta\) to the horizontal, as in Figure 11.1. 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*}
and
\begin{equation*}
\tan\phi = \frac{v_0\sin\theta - g\,t}{v_0\cos\theta}\text{.}
\end{equation*}
Solution.
Take the origin at \(O\) with horizontal and vertical axes. The coordinates of \(P\) at time \(t\) are
\begin{equation*}
x = v_0\cos\theta\;t\text{,} \qquad
y = v_0\sin\theta\;t - \tfrac{1}{2}g\,t^2\text{.}
\end{equation*}
The horizontal and vertical components of the velocity are the time derivatives of these, so
\begin{equation*}
v\cos\phi = \frac{dx}{dt} = v_0\cos\theta\text{,} \qquad
v\sin\phi = \frac{dy}{dt} = v_0\sin\theta - g\,t\text{.}
\end{equation*}
Squaring and adding eliminates \(\phi\text{:}\)
\begin{equation*}
v^2 = v_0^2\cos^2\theta + \bigl(v_0\sin\theta - g\,t\bigr)^2
= v_0^2 - 2\,v_0\,g\,t\sin\theta + g^2 t^2\text{,}
\end{equation*}
which gives the speed at time \(t\text{.}\) Dividing the vertical component by the horizontal component gives the direction,
\begin{equation*}
\tan\phi = \frac{v_0\sin\theta - g\,t}{v_0\cos\theta}\text{,}
\end{equation*}
that is, \(\phi = \arctan\!\left(\dfrac{v_0\sin\theta - g\,t}{v_0\cos\theta}\right)\text{.}\)
