Activity 1.3.1. Projectiles on an Inclined Plane.
A particle is launched from a point \(O\) on an inclined plane and travels in the vertical plane that contains the line of greatest slope through \(O\text{.}\) The plane makes an angle \(\beta\) with the horizontal, and the particle leaves \(O\) with initial speed \(v_0\) at an elevation \(\theta\) to the horizontal. Taking the origin at \(O\) with horizontal and vertical axes and neglecting air resistance, the trajectory is the projectile parabola
\begin{equation*}
y = x\tan\theta - \frac{g\,x^2}{2\,v_0^2\cos^2\theta}\text{.}
\end{equation*}
Let \(P\) be the point where the particle meets the plane again, and let \(R = OP\) be the range measured along the plane, so that \(P = \bigl(R\cos\beta,\ R\sin\beta\bigr)\text{.}\) Work through the tasks below to find the range on the plane, the maximum range for a given \(v_0\text{,}\) and the time of flight, and then to read off the corresponding results when the particle is fired down the plane.
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(a)
Since \(P=\bigl(R\cos\beta,\ R\sin\beta\bigr)\) lies on the trajectory, substitute these coordinates into the parabola and cancel a factor of \(R\) to show that
\begin{equation*}
\sin\beta = \cos\beta\,\tan\theta
- \frac{g\,R\,\cos^2\beta}{2\,v_0^2\cos^2\theta}\text{.}
\end{equation*}
Solution.
Putting \(x=R\cos\beta\) and \(y=R\sin\beta\) into \(y = x\tan\theta - \dfrac{g x^2}{2 v_0^2\cos^2\theta}\) gives
\begin{equation*}
R\sin\beta = R\cos\beta\,\tan\theta
- \frac{g\,R^2\cos^2\beta}{2\,v_0^2\cos^2\theta}\text{.}
\end{equation*}
Dividing through by \(R\) (the particle is not at \(O\)) yields the stated relation.
(b)
Rearrange the relation from the previous task to isolate \(R\text{,}\) and use \(\sin\theta\cos\beta - \cos\theta\sin\beta = \sin(\theta-\beta)\) to obtain the range up the plane
\begin{equation*}
R = \frac{2\,v_0^2\,\sin(\theta-\beta)\cos\theta}{g\cos^2\beta}\text{.}
\end{equation*}
Solution.
Collect the \(R\)-term:
\begin{equation*}
\frac{g\,R\,\cos^2\beta}{2\,v_0^2\cos^2\theta}
= \cos\beta\,\tan\theta - \sin\beta
= \frac{\sin\theta\cos\beta - \cos\theta\sin\beta}{\cos\theta}
= \frac{\sin(\theta-\beta)}{\cos\theta}\text{.}
\end{equation*}
Solving for \(R\text{,}\)
\begin{equation*}
R = \frac{2\,v_0^2\cos^2\theta}{g\cos^2\beta}\cdot
\frac{\sin(\theta-\beta)}{\cos\theta}
= \frac{2\,v_0^2\,\sin(\theta-\beta)\cos\theta}{g\cos^2\beta}\text{.}
\end{equation*}
(c)
Using the identity \(2\sin(\theta-\beta)\cos\theta = \sin(2\theta-\beta) - \sin\beta\text{,}\) rewrite the range as
\begin{equation*}
R = \frac{v_0^2}{g\cos^2\beta}
\bigl[\sin(2\theta-\beta) - \sin\beta\bigr]\text{.}
\end{equation*}
For a fixed launch speed \(v_0\text{,}\) deduce the elevation \(\theta\) that maximizes \(R\text{,}\) and show that the maximum range up the plane is
\begin{equation*}
R_{\max} = \frac{v_0^2}{g\,(1+\sin\beta)}\text{.}
\end{equation*}
Hint.
With \(v_0\text{,}\) \(g\text{,}\) and \(\beta\) held constant, only \(\sin(2\theta-\beta)\) varies, and it is largest when it equals \(1\text{.}\) Afterwards use \(\cos^2\beta = (1-\sin\beta)(1+\sin\beta)\text{.}\)
Solution.
\(R\) is greatest when \(\sin(2\theta-\beta)=1\text{,}\) i.e. when \(2\theta-\beta = \dfrac{\pi}{2}\text{,}\) giving the optimal elevation
\begin{equation*}
\theta = \frac{\pi}{4} + \frac{\beta}{2}\text{.}
\end{equation*}
At this value,
\begin{equation*}
R_{\max} = \frac{v_0^2}{g\cos^2\beta}\,(1-\sin\beta)
= \frac{v_0^2\,(1-\sin\beta)}{g\,(1-\sin\beta)(1+\sin\beta)}
= \frac{v_0^2}{g\,(1+\sin\beta)}\text{.}
\end{equation*}
(d)
The horizontal distance \(OM\) to the foot of \(P\) is covered at the constant horizontal speed \(v_0\cos\theta\text{.}\) Using \(OM = R\cos\beta\text{,}\) show that the time of flight is
\begin{equation*}
T = \frac{2\,v_0\,\sin(\theta-\beta)}{g\cos\beta}\text{.}
\end{equation*}
Solution.
From \(R\cos\beta = v_0\cos\theta\cdot T\text{,}\)
\begin{equation*}
T = \frac{R\cos\beta}{v_0\cos\theta}
= \frac{\cos\beta}{v_0\cos\theta}\cdot
\frac{2\,v_0^2\,\sin(\theta-\beta)\cos\theta}{g\cos^2\beta}
= \frac{2\,v_0\,\sin(\theta-\beta)}{g\cos\beta}\text{.}
\end{equation*}
(e)
When the particle is projected down the plane, the geometry is the same with \(\beta\) replaced by \(-\beta\text{.}\) Write down the range, maximum range, and time of flight down the inclined plane.
Answer.
Replacing \(\beta \to -\beta\text{:}\)
\begin{equation*}
R_{\text{down}} =
\frac{2\,v_0^2\,\sin(\theta+\beta)\cos\theta}{g\cos^2\beta}\text{,}
\qquad
R_{\max,\text{down}} = \frac{v_0^2}{g\,(1-\sin\beta)}\text{,}
\end{equation*}
\begin{equation*}
T_{\text{down}} = \frac{2\,v_0\,\sin(\theta+\beta)}{g\cos\beta}\text{.}
\end{equation*}
The maximum range up the plane occurs when \(2\theta - \beta = \dfrac{\pi}{2}\text{,}\) which rearranges to \(\theta - \beta = \dfrac{\pi}{2} - \theta\text{.}\) The left side is the angle the launch direction makes above the incline, and the right side is the angle it makes below the vertical. Since they are equal, the direction of projection for maximum range up an inclined plane bisects the angle between the upward vertical through \(O\) and the line of greatest slope.
