Projectile motion

In a nutshell

Projectiles have components of vertical and horizontal motion which must be analysed individually, as they have no effect on each other. The horizontal velocity of a projectile is constant, whilst the vertical will change as the object accelerates due to gravity.

Equations

DESCRIPTION	EQUATION
Velocity	$v = \sqrt{v_x^2 + v_y^2}$
Angle from horizontal	$\theta = \tan^{-1}\left(\dfrac{v_x}{v_y}\right)$

Constants

CONSTANT	SYMBOL	VALUE
$acceleration \ due \ to \ gravity$	$g$	- $9.81 \ ms^{-2}$

Variables

QUANTITY NAME	SYMBOL	DERIVED UNIT	SI UNIT
$x \ component \ of \ velocity$	$v_x$	$ms^{-1}$	$ms^{-1}$
$y \ component \ of \ velocity$	$v_y$	$ms^{-1}$	$ms^{-1}$
$angle \ from \ the \ horizontal$	$\theta$	$\degree$	$rad$
$displacement$	$s$	$m$	$m$
$time \ taken$	$t$	$s$	$s$

Projectiles

A projectile is an object that is thrown at an angle to the horizontal, and therefore has motion in both the horizontal and vertical plane. The horizontal and vertical motions are independent of each other, so they must be analysed on their own.

However, time is interchangeable between the directions of motion.

In projectile motion, air resistance is assumed to be negligible:

the vertical velocity changes due to the acceleration due to gravity, $g$ .
the vertical components can be calculated using equations of motion using $a=g$ .
the horizontal velocity is constant.

Physics; Motion; KS5 Year 12; Projectile motion

1.	Vertical component for velocity will increase with time
2.	Horizontal component for velocity will remain constant
3.	Projectile path
4.	Horizontal displacement
5.	Vertical displacement

Horizontal velocity

The horizontal velocity remains constant as there are no forces acting in the horizontal direction. The horizontal velocity can therefore be analysed using the equation:

$v_x = \dfrac{s_x}{t}$

The component of initial horizontal velocity is:

$u_x = u \cos(\theta)$

Vertical velocity

The vertical velocity changes due to the acceleration of free fall $g$ . The suvat equations of motion can be used to analyse the vertical motion. The initial velocity is:

$u_y = u \sin(\theta)$

Sometimes projectile problems require the journey to be split into two, the first half from launch to max height and then the second half from max height to ground.

Note: The direction of motion must be defined for the whole question.

Resolving velocities

The path of a projectile is commonly curved, as the vertical velocity increases whilst the horizontal velocity remains constant. The magnitude of the velocity can be calculated by using the horizontal and vertical components of velocity:

$v = \sqrt{v_x^2 + v_y^2}$

The angle made by the velocity to the horizontal is given by:

$\theta = \tan^{-1}\left(\dfrac{v_x}{v_y}\right)$

Example

An object is launched at a velocity of $40 \ ms^{-1}$ at an angle of $50 \ \degree$ above the horizontal. Calculate the maximum height reached.

Firstly, considering the vertical components, state the variables in the question:

$s_y= \ ?\newline u=40 \ ms^{−1}\newline v_y= 0 \ ms^{-1}\newline a_y= -9.81 \ ms^{-1}\newline t= \boxtimes$
$\theta = 50 \ \degree$

Next find the vertical velocity component:

$u_y = u \sin(\theta)\newline \\[0.1in] u_y = 40 \sin(50) = 30.6 \ ms^{-1}$

State the equation needed:

$v_y^2 = u_y^2 + 2a_ys_y$

Rearrange for $s$ :

$s_y = \dfrac{v_y^2 - u_y^2}{2a_y} \\[0.1in]\newline s_y = \dfrac{-u_y^2}{-2g}$

Substitute in and solve (negative sign disappears as they cancel):

$s = \dfrac{(30.6)^2}{2 \times 9.81} \\[0.1in]\newline s= 47 \ m\ (2sf)$

The maximum height reached was $\underline{47 \ m}$ .