📂Classical Mechanics

Parabolic Motion: Horizontal Range and Maximum Height Angle

Definition^{1 2}

An object launched with angle $\alpha$ and initial speed $v_{0}$ performs a motion known as parabolic motion.

Description

It’s also referred to as projectile motion.

Typically, external forces like air resistance are ignored, so the motion is uniform in the horizontal direction and free fall in the vertical direction.

Analysis

Motion in the $x$ direction (horizontal) is independent of gravity, while motion in the $y$ direction (vertical) is affected by gravity.

$$ \begin{align*} x &= (v_o\cos\alpha)t && & y &= -\frac{1}{2}gt^{2}+(v_{0}\sin\alpha)t \\ v_{x} &= v_{0}\cos\alpha && & v_{y} &= -gt+v_{0}\sin\alpha \\ a_{x} &= 0 && & a_{y} &= -g \\ F_{x} &= 0 && & F_{y} &= -mg \end{align*} $$

Let’s take two equations concerning positions in vertical and horizontal directions.

$$ \begin{align} x &= (v_{0}\cos\alpha)t \\ y &= -\frac{1}{2}gt^{2}+(v_{0}\sin\alpha)t \end{align} $$

By arranging for the commonly included $t$, we get an equation purely in terms of $x, y, \alpha$. In other words, we can know information about the horizontal distance and vertical height of the parabolic motion as per the angle. Arranging $(1)$ in terms of $t$ gives us $t=\dfrac{x}{v_{0}\cos\alpha}$, and by substituting in $(2)$ and arranging it as a quadratic equation regarding $x$,

$$ \begin{align} y &= -\frac{1}{2}g\left(\frac{x}{v_{0}\cos\alpha}\right)^{2}+(v_{0}\sin\alpha)\left(\frac{x}{v_{0}\cos\alpha}\right) \nonumber \\ &= -\frac{g}{2{v_{0}}^{2}\cos^{2}\alpha}x^{2}+\left(\frac{\sin\alpha}{\cos\alpha}\right)x \nonumber \\ &= -\frac{g}{2{v_{0}}^{2}\cos^{2}\alpha}\left(x^{2}-\frac{2{v_{0}}^{2}\sin\alpha \cos\alpha}{g}x\right) \\ &= -\frac{g}{2{v_{0}}^{2}\cos^{2}\alpha}\left(x^{2}-\frac{2{v_{0}}^{2}\sin\alpha \cos\alpha}{g}x+\frac{{v_{0}}^4\sin^{2}\alpha \cos^{2}\alpha}{g^{2}}\right) +\frac{{v_{0}}^{2}\sin^{2}\alpha}{2g} \nonumber \\ &= -\frac{g}{2{v_{0}}^{2}\cos ^{2}\alpha}\left(x-\frac{{v_{0}}^{2}\sin\alpha \cos\alpha}{g}\right)^{2} +\frac{{v_{0}}^{2}\sin^{2}\alpha}{2g} \nonumber \end{align} $$

Maximum Height

From the above equation, we can easily see that the vertex of the parabolic motion graph is at $\left(\dfrac{{v_{0}}^{2}\sin\alpha \cos\alpha}{g}, \dfrac{{v_{0}}^{2}\sin^{2}\alpha}{2g} \right)$. Therefore, based on the launch angle $\alpha$ and initial speed $v_{0}$, the maximum height is

$$ y = \dfrac{{v_{0}}^{2}\sin^{2}\alpha}{2g} $$

When the angle for the maximum height is (?) at $\alpha = 90^{\circ} = \frac{\pi}{2}$,

$$ y = \dfrac{v_{0}^{2}}{2g} $$

Horizontal Range

A root other than $0$ of the parabolic motion graph is the horizontal range. Based on the $(3)$ formula,

$$ \frac{g}{2{v_{0}}^{2}\cos^{2}\alpha}\left(x^{2}-\frac{2{v_{0}}^{2}\sin\alpha \cos\alpha}{g}x\right) = \frac{g}{2{v_{0}}^{2}\cos^{2}\alpha} x \left(x-\frac{2{v_{0}}^{2}\sin\alpha \cos\alpha}{g}\right) $$

Hence, the horizontal range can be described as follows.

$$ \begin{equation} x = \dfrac{2{v_{0}}^{2}\sin\alpha \cos\alpha}{g} = \dfrac{{v_{0}}^{2}\sin2\alpha}{g} \end{equation} $$

Let’s take the two equations we arranged regarding $x$ and $y$. Both equations include time $t$. Our question does not concern time. In other words, by arranging one equation for $t$ and substituting it into the other, we can answer the question we were interested in.

Flight Time

If we regard the reach distance as $R = \dfrac{{v_{0}}^{2}\sin2\alpha}{g}$, then

$$ t = \dfrac{{v_{0}}^{2}\sin2\alpha}{g v_{0}\cos\alpha} = \dfrac{R}{v_{0}\cos\alpha} $$

Vertical Motion

If $\alpha = 90^{\circ} = \frac{\pi}{2}$, it describes an object moving in vertical direction.

• Horizontal Range: $x = \dfrac{{v_{0}}^{2}\sin \pi}{g} = 0$

• Vertical Height: $y = \dfrac{v_{0}^{2}}{2g}$

• Flight Time: $t = \dfrac{v_{0}}{g}$

Also, an object falling from the maximum height undergoes [free fall].