What is Projectile Motion: Definition, Formula Derivation, Equations, Examples, Notes

The movement of a projectile in air, acted upon by gravity as the only force, is known as projectile motion. As an example of two-dimensional motion where acceleration is constant, it combines two simultaneous motions that are mutually perpendicular and independent of each other.  

In a graph, you will see horizontal and vertical motion when trying to decipher projectile motion. 

Horizontal motion alone = straight line sideways
Vertical motion alone = straight line up and down
Horizontal motion + Vertical motion = Parabolic path (the curved trajectory)

The NCERT book describes projectile motion as 
"An object that is in flight after being thrown or projected is called a projectile. Such a projectile might be a football, a cricket ball, a baseball or any other object. The motion of a projectile may be thought of as the result of two separate, simultaneously occurring components of motions. One component is along a horizontal direction without any acceleration and the other along the vertical direction with constant acceleration due to the force of gravity."

Galileo stated in his Dialogue on the Great World Systems, 1632, that two perpendicular directions of motion are always independent of each other. So in physics, the vector quantity directed along a direction will not be affected by a vector perpendicular to it.

• Let's say a projectile that is thrown with a velocity u is making an angle theta with the horizontal.
• The initial velocity gets resolved into components within a coordinate system. The horizontal direction, then, is taken as x-axis, while vertical direction as y-axis and point of projection as the origin.
• We can also consider this projectile motion as the combination of horizontal and vertical motion. So, Horizontal direction and Vertical Direction can be seen as

(a) Initial velocity $u_x=u\mathrm{c}\mathrm{o}\mathrm{s}\theta$ for horizontal

Initial velocity ${\mathrm{u}}_{\mathrm{y}}=\mathrm{u}\mathrm{s}\mathrm{i}\mathrm{n}\theta$ for vertical

(b) Acceleration $a_x=0$ for horizonal 

Acceleration ${\mathrm{a}}_{\mathrm{y}}=\mathrm{g}$ for vertical

(c) Velocity after time $\mathrm{t},{\mathrm{v}}_{\mathrm{x}}=\mathrm{u}\mathrm{c}\mathrm{o}\mathrm{s}\theta$ for horizontal

Velocity after time $\mathrm{t},{\mathrm{v}}_{\mathrm{y}}=\mathrm{u}\mathrm{s}\mathrm{i}\mathrm{n}\theta -\mathrm{g}\mathrm{t}$ for vertical

Time of Flight 

The displacement then along the vertical direction will be zero for the entire flight.

Hence, along vertical direction net displacement $=0$

$\Rightarrow (u\mathrm{s}\mathrm{i}\mathrm{n}\theta )T-\frac{1}{2}g{T}^2=0\mathrm{}\Rightarrow \mathrm{}T=\frac{2u\mathrm{s}\mathrm{i}\mathrm{n}\theta }{g}$

Horizontal Range

$\begin{array}{rr}& R=u_x\cdot T\mathrm{}\Rightarrow \mathrm{}R=u\mathrm{c}\mathrm{o}\mathrm{s}\theta \cdot \frac{2u\mathrm{s}\mathrm{i}\mathrm{n}\theta }{g}\\ & R=\frac{u^2\mathrm{s}\mathrm{i}\mathrm{n}2\theta }{g}\end{array}$

Maximum Height

The trajectory's highest point sees the object moving in horizontal direction, so the vertical component of velocity will be zero.

Using $3^{\text{rd}}$ equation of motion i.e. $v^2=u^2+2$ as we have for vertical direction $0=u^2{\mathrm{s}\mathrm{i}\mathrm{n}}^2\theta -2gH\mathrm{}\Rightarrow \mathrm{}H=\frac{u^2{\mathrm{s}\mathrm{i}\mathrm{n}}^2\theta }{2g}$

Resultant Velocity

$\vec{v}=v_x\hat{i}+v_y\hat{j}=u\mathrm{c}\mathrm{o}\mathrm{s}\theta \hat{i}+(u\mathrm{s}\mathrm{i}\mathrm{n}\theta -gt)\hat{j}$

Where, $\left|\stackrel{\rightharpoonup }{\mathrm{v}}\right|=\sqrt[]{{\mathrm{u}}^2{\mathrm{c}\mathrm{o}\mathrm{s}}^2\theta +(\mathrm{u}\mathrm{s}\mathrm{i}\mathrm{n}\theta -\mathrm{g}\mathrm{t}{)}^2}$ and $\mathrm{t}\mathrm{a}\mathrm{n}\alpha ={\mathrm{v}}_{\mathrm{y}}/{\mathrm{v}}_{\mathrm{x}}$ .
Also, $v\mathrm{c}\mathrm{o}\mathrm{s}\alpha =u\mathrm{c}\mathrm{o}\mathrm{s}\theta \mathrm{}\Rightarrow \mathrm{}v=\frac{u\mathrm{c}\mathrm{o}\mathrm{s}\theta }{\mathrm{c}\mathrm{o}\mathrm{s}\alpha }$

General Result

1. For maximum range $\theta ={45}^{\circ }$

$R_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{u^2}{g}\mathrm{}\Rightarrow \mathrm{}{H}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{R_{\mathrm{m}\mathrm{a}\mathrm{x}}}{2}$

2. We find the same range for two projection angles $\alpha$ and ( $90-\alpha$ ). However, in these two cases, the maximum height the particles go up to would be different.
This is because, $R=\frac{u^2\mathrm{s}\mathrm{i}\mathrm{n}2\theta }{g}$ , and $\mathrm{s}\mathrm{i}\mathrm{n}2(90-\alpha )=\mathrm{s}\mathrm{i}\mathrm{n}180-2\alpha =\mathrm{s}\mathrm{i}\mathrm{n}2\alpha$

3. If $R=H$
i.e. $\frac{u^2\mathrm{s}\mathrm{i}\mathrm{n}2\theta }{g}=\frac{u^2{\mathrm{s}\mathrm{i}\mathrm{n}}^2\theta }{2g}\mathrm{}\Rightarrow \mathrm{}\mathrm{t}\mathrm{a}\mathrm{n}\theta =4$

4. We can also express range as $R=\frac{u^2\mathrm{s}\mathrm{i}\mathrm{n}2\theta }{g}=\frac{2u\mathrm{s}\mathrm{i}\mathrm{n}\theta \cdot u\mathrm{c}\mathrm{o}\mathrm{s}\theta }{g}=\frac{2u_x{u}_y}{g}$

The path a particle or projectile follows during motion is its Trajectory. So the equation of trajectory is the relation between instantaneous coordinates (x and y), of the particle. When we consider the horizontal direction,

$x=u_{x.t}$

$\begin{array}{c}x=ucos\theta .t\#\left(1\right)\end{array}$

For vertical direction : $\begin{array}{rr}y& =u_y\cdot t-1/2g{t}^2\end{array}$

Eliminating ' t ' from equation (1) & (2)

$y=u\mathrm{s}\mathrm{i}\mathrm{n}\theta \cdot \frac{x}{u\mathrm{c}\mathrm{o}\mathrm{s}\theta }-\frac{1}{2}g{\left(\frac{x}{u\mathrm{c}\mathrm{o}\mathrm{s}\theta }\right)}^2\Rightarrow \mathrm{}y=x\mathrm{t}\mathrm{a}\mathrm{n}\theta -\frac{g{x}^2}{2u^2{\mathrm{c}\mathrm{o}\mathrm{s}}^2\theta }$

This is the parabola equation and is known as the trajectory equation of projectile motion.

Here are some other forms of the trajectory/parabola equation. 

$y=x\mathrm{t}\mathrm{a}\mathrm{n}\theta -\frac{g{x}^2\left(1+{\mathrm{t}\mathrm{a}\mathrm{n}}^2\theta \right)}{2u^2}$

ey $y=x\mathrm{t}\mathrm{a}\mathrm{n}\theta -\frac{g{x}^2}{2u^2{\mathrm{c}\mathrm{o}\mathrm{s}}^2\theta }$
$\Rightarrow y=x\mathrm{t}\mathrm{a}\mathrm{n}\theta \left[1-\frac{gx}{2u^2{\mathrm{c}\mathrm{o}\mathrm{s}}^2\theta \mathrm{t}\mathrm{a}\mathrm{n}\theta }\right]$

$\Rightarrow y=x\mathrm{t}\mathrm{a}\mathrm{n}\theta \left[1-\frac{gx}{2u^2\mathrm{s}\mathrm{i}\mathrm{n}\theta \mathrm{c}\mathrm{o}\mathrm{s}\theta }\right]$

$\Rightarrow y=x\mathrm{t}\mathrm{a}\mathrm{n}\theta \left[1-\frac{x}{R}\right]$

Consider a projectile thrown from point O  at some height  from the ground with a velocity u. Let's review projectile motion's characteristics, only by resolving motion along both directions: horizontal and vertical. 

Horizontal direction

(i) Initial velocity $u_x=u$

(ii) Acceleration $a_x=0$

Vertical direction

(i) Initial velocity $u_y=0$

(ii) Acceleration ${\mathrm{a}}_{\mathrm{y}}=\mathrm{g}$ (downward)

Here, it is equal to the time the projectile takes to return to the ground. So, from the equation of motion $S=ut+\frac{1}{2}a{t}^2$ , alongside the vertical direction, we get $-\mathrm{h}={\mathrm{u}}_{\mathrm{y}}\mathrm{t}+\frac{1}{2}(-\mathrm{g}){\mathrm{t}}^2\Rightarrow \mathrm{h}=\frac{1}{2}\mathrm{g}{\mathrm{t}}^2\mathrm{}\Rightarrow \mathrm{}\mathrm{t}=\sqrt[]{\frac{2\mathrm{h}}{\mathrm{g}}}$