A type of parametric equation you may have already seen is the equation of motion for a projectile. Here the parameter is time, and \((x(t), y(t))\) gives the position of the projectile at time \(t\text{.}\)
Recall two basic facts about motion in one dimension from your introductory physics course. For motion along a straight line at constant speed \(v\text{,}\) the position is
\begin{equation*}
x(t) = v\, t + x_0.
\end{equation*}
For motion with uniform acceleration \(a\text{,}\) the position is
\begin{equation*}
x(t) = \frac{1}{2} a t^2 + v_0 t + x_0.
\end{equation*}
Now consider an object moving under the influence of gravity alone, ignoring air resistance. No horizontal force acts on it, so it moves at constant speed in the \(x\)-direction; in the \(y\)-direction it moves with constant acceleration \(g\text{,}\) directed downward. Resolving the initial velocity into its horizontal component \(v_0 \cos(\theta)\) and vertical component \(v_0 \sin(\theta)\text{,}\) and applying the two facts above, gives the equations of motion for a projectile:
\begin{align}
x(t) \amp = v_0 \cos(\theta)\, t + x_0,\tag{1.7}\\
y(t) \amp = - \frac{1}{2} g t^2 + v_0 \sin(\theta)\, t + y_0,\tag{1.8}
\end{align}
where \(v_0\) is the initial speed, \(\theta\) is the launch angle, and \(g\) is the acceleration due to gravity. For simplicity, we take the initial position \((x_0, y_0)\) to be the origin \((0,0)\text{.}\)
To find the trajectory of the projectile, we can solve for \(t\) and substitute into the \(y(t)\) equation (1.8). Solving for \(t\) in the \(x(t)\) equation (1.7) gives
\begin{align}
t \amp = \frac{x}{v_0 \cos(\theta)}.\tag{1.9}
\end{align}
As expected, the trajectory equation (1.10) shows that the trajectory of a projectile is a parabola. Also, note that differentiating the equations of motion gives the velocity of the projectile: the horizontal component \(v_x = v_0\cos(\theta)\) stays constant, since nothing accelerates the projectile horizontally, while the vertical component \(v_y = v_0\sin(\theta) - g t\) decreases steadily under gravity, so it is positive while the projectile rises, zero at the apex, and negative as it falls. Figure 1.17 shows these components at five instants along the path.
Figure1.17.The velocity of a projectile at five instants along its parabolic path. The horizontal component \(\vec v_x\) stays constant while the vertical component \(\vec v_y\) changes under the constant downward acceleration \(\vec g\text{;}\) at the apex the vertical component is zero.
Figure 1.18 animates this motion: the projectile follows the parabolic path while its horizontal velocity stays constant and its vertical velocity is steadily turned around by gravity.
Figure1.18.A projectile launched from the origin traces out its parabolic trajectory. The horizontal velocity component \(v_x\) is constant, while the vertical component \(v_y\) shrinks to zero at the apex and then grows in the downward direction as the projectile falls.
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.
Figure1.20.A particle projected up the plane with speed \(v_0\) at elevation \(\theta\text{,}\) striking the incline (angle \(\beta\)) at \(P\) after travelling a range \(R=OP\) along the slope.
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
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*}
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*}
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.
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.
