We consider an example to understand gravitational acceleration when there is no air resistance. Two objects, one is a metal ball and other is a small piece of paper fall from a particular height. Do they reach the ground at the same time? Your answer may be no but the correct answer is yes if there is no air resistance. The air resistance causes the piece of paper fall slowly.

If you neglect the air resistance or if the air resistance is zero, both objects reach the ground at the same time. So in conclusion the acceleration due to gravity or gravitational acceleration is independent of mass, that is all objects have the same acceleration.

The gravitational acceleration is denoted by \(g\) whose value on the Earth's surface is \(9.8\text{m/}{{\text{s}}^{\text{2}}}\) in two significant figures. It's value is not the same in all places on the Earth but we consider this value for now.

## What is projectile motion?

A thrown object is called projectile and its motion under the action of gravitational force is called projectile motion. When you through something, the thrown object goes along a curve and falls back to the ground at some distance apart.The projectile motion entirely depends on the acceleration due to gravity, or the gravitational field that continuously provides a gravitational acceleration or any other field that provides the constant acceleration. But for now we are talking about the projectile motion on the Earth's surface considering that there is no interfering air resistance.

There are two kinds of motion associated with projectile motion; one is horizontal and another is vertical. So, it is two dimensional motion as both horizontal and vertical coordinates change during the motion.

We represent the x-coordinate to be the horizontal and the y-coordinate to be the vertical coordinate in a vertical xy-plane while describing the projectile motion. Now we are ready to launch a projectile. In Figure 1 the projectile is launched at the origin of the coordinate system whose initial launch velocity is \({{\vec{v}}_{i}}\).

The projectile is launched (thrown) at the initial angle \(\theta\). The initial velocity can be resolved into x and y-components. The magnitude of the x-component of \({{\vec{v}}_{i}}\) is \({{v}_{ix}}={{v}_{i}}\cos \theta \) and the magnitude of the y-component of of \({{\vec{v}}_{i}}\) is \({{v}_{iy}}={{v}_{i}}\sin \theta \).

Projectile motion is the combination of horizontal and vertical motion. The speed of the horizontal motion remains the same but it changes in vertical motion. So \({{v}_{x}}={{v}_{ix}}\) where \({{v}_{x}}\) is the horizontal velocity (the subscript \(x\) reminds the velocity along x-axis).

The acceleration in vertical motion is \(-g\). The negative sign before \(g\) is due to the vertical motion, that is, the projectile moves upwards and falls downwards.

The y-velocity (velocity along y-axis) \({{v}_{y}}\) at any point on the curve in time \(t\) is, \[{{{v}_{y}}={{v}_{iy}}-gt=({{v}_{i}}\sin \theta )-gt \tag{1} \label{1}}\]But in projectile motion the velocity in horizontal motion remains the same, that is the velocity along x-axis remains the same, so, \[{{v}_{x}}={{v}_{ix}}+{{a}_{x}}t={{v}_{ix}}={{v}_{i}}\cos \theta \tag{2} \label{2}\] The magnitude and the direction of velocity of the projectile at any point on the curve are, \(v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}\) and \(\phi =\arctan \left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)\).

The next approach is to find the equation of the trajectory of the projectile. For this we find the position of the projectile at any point on its path. Here we can use equations of straight line motion with constant acceleration to find the x and y coordinates of the projectile at any point, so, \[\begin{align*} x&={{v}_{ix}}t={{v}_{x}}t=({{v}_{i}}\cos \theta )t \tag{3} \label{3}\\ y&={{v}_{iy}}t-\frac{1}{2}g{{t}^{2}}=({{v}_{i}}\sin \theta )t-\frac{1}{2}g{{t}^{2}} \tag{4} \label{4} \end{align*}\]Now putting the value of \(t=\frac{x}{{{v}_{i}}\cos \theta }\) from Eq. \eqref{3} into Eq. \eqref{4} and solving you'll get, \[\begin{align*} y&=({{v}_{i}}\sin \theta )\left( \frac{x}{{{v}_{i}}\cos \theta } \right)-\frac{1}{2}g{{\left( \frac{x}{{{v}_{i}}\cos \theta } \right)}^{2}} \\ \text{or,}\quad y&=(\tan \theta )x-\left( \frac{g}{2{{v}_{i}}^{2}{{\cos }^{2}}\theta } \right){{x}^{2}} \tag{5} \label{5} \end{align*}\] Note that the angle \(\theta\) is the initial angle at which the projectile is launched. The equation Eq. \eqref{5} is the equation of the trajectory of the projectile and this is the equation of a parabola. Thus, we can conclude that the equation of a projectile is parabolic in nature.

### What is maximum height and horizontal range of a projectile?

At the maximum height \({{v}_{y}}\) is zero, so from Eq. \eqref{1} when \({{v}_{y}}=0\), we have \(t=\frac{{{v}_{i}}\sin \theta }{g}\) which is the time when the projectile reaches the maximum height. Now replacing \(t\) in Eq. \eqref{4} by \(t=\frac{{{v}_{i}}\sin \theta }{g}\) we get the maximum height \(h\), so \({{y}_{\max }}=h\) is \[\begin{align*} {{y}_{\max }}&=h=({{v}_{i}}\sin \theta )\left( \frac{{{v}_{i}}\sin \theta }{g} \right)-\frac{1}{2}g{{\left( \frac{{{v}_{i}}\sin \theta }{g} \right)}^{2}} \\ \text{or,}\quad h&=\frac{{{v}_{i}}^{2}{{\sin }^{2}}\theta }{2g} \tag{6} \label{6} \end{align*}\] The total distance between the launch point and the point where the projectile hits the ground is the maximum horizontal range \(R\). Now we find the time in which the projectile hits the ground. When the projectile hits the ground \(y = 0\) in Eq. \eqref{4} so, \[\begin{align*} ({{v}_{i}}\sin \theta )t-\frac{1}{2}g{{t}^{2}}&=0 \\ \left( {{v}_{i}}\sin \theta -\frac{1}{2}gt \right)t&=0 \\ \text{Either,}\ \left( {{v}_{i}}\sin \theta -\frac{1}{2}gt \right)&=0\Rightarrow t=\frac{2{{v}_{i}}\sin \theta }{g} \\ \text{or,}\ t&=0 \\ \end{align*}\]The time \(t=0\) is the initial time when the projectile was lunched at the origin but \(t=\frac{2{{v}_{i}}\sin \theta }{g}\) is the time when the projectile hit the ground.

In both cases the height of the projectile \(y\) is zero. To find the maximum horizontal range we use \(t=\frac{2{{v}_{i}}\sin \theta }{g}\) and put this value in Eq. \eqref{3}. Using Eq. \eqref{3} the maximum horizontal range \({{x}_{\max }}=R\) is,

\[\begin{align*} & {{x}_{\max }}=R={{v}_{i}}\cos \theta \left( \frac{2{{v}_{i}}\sin \theta }{g} \right)=\frac{2{{v}_{i}}^{2}\sin \theta \cos \theta }{g} \\ & \text{or,}\quad R=\frac{{{v}_{i}}^{2}\sin 2\theta }{g} \tag{7} \label{7} \end{align*}\]

The time the projectile takes to the reach the ground is two times the time it takes to reach the maximum height.

The above derivation and the nature of how a projectile motion takes place lead us to understand that the two motions a projectile has are completely independent with each other. It is the same thing as one motion does not know the existence of the other motion and vice versa. And we can simply use the equations of motion (kinematic equations) for solving the complicated looking problem easily (equation of the trajectory of the projectile).