Horizontal projection

In a nutshell

Projectile motion involves the study of how a particle moves under the influence of gravity. When considering the motion of a particle in 2D, the vertical and horizontal components are independent of each other, and therefore the calculations for vertical and horizontal movement can be considered separately.

Horizontal projection

When considering the movement of a particle projected horizontally, consider the horizontal and vertical motion separately. Assume air resistance is zero and ignore any rotational effects of the particle.

Horizontal motion

There is no acceleration in the horizontal direction as the only acceleration is vertical, (acceleration due to gravity). Therefore, the horizontal component of a projectile is considered to have constant velocity. The equation $\bold s= \bold v t$ is used.

Vertical motion

When projected horizontally, the initial vertical component of the velocity is $0 \ m.s^{-1}$ . Acceleration due to gravity is $9.8 \ m.s^{-2}$ , and so the formulae for constant acceleration can be used in the vertical direction.

Equations

description

equation

Horizontal motion is modelled as having

constant velocity

(a=0)

.

s=vt

Vertical motion is modelled as having

constant acceleration due to gravity,

so the constant acceleration formulae can be used.

\begin{aligned}v&=u+at \\ \\s&=\dfrac{(u+v)}{2}t \\\\s&=ut+\dfrac{1}{2}at^2 \\\\s&=vt-\dfrac{1}{2}at^2 \\\\v^2&=u^2+2as\end{aligned}

Variable definitions

Quantity name	symbol	unit name	unit
$displacement$	$s$	$metre$	$m$
$initial \ velocity$	$u$	$metres \ per \ second$	$m.s^{-1}$
$final \ velocity$	$v$	$metres \ per \ second$	$m.s^{-1}$
$acceleration$	$a$	$metres \ per \ second \ squared$	$m.s^{-2}$
$time$	$t$	$seconds$	$s$
$acceleration \ due \ to \ gravity$	$g$ $=9.8$	$metres \ per \ second \ squared$	$m.s^{-2}$

Example

A particle is projected horizontally at $20\ m.s^{-1}$  from $80\ m$ above the ground. Find the time taken to reach the ground and the horizontal range.

First draw a sketch to show the motion of the particle:

Maths; Projectiles; KS5 Year 13; Horizontal projection

For this question, take vertically up to be positive, and horizontally right to be positive.

To calculate the time taken, use the vertical motion. First state all the known vertical values:

 $s=-80 \qquad u = 0 \qquad v= ? \qquad a=-9.8 \qquad t=?$ 

As $v$ is unknown, use $s=ut+\dfrac 1 2 at^2$ to calculate $t$ :

$\begin{aligned}-80&=0+\cfrac{1}{2} \times -9.8 \times t^2\\-80&=-4.9t^2\\ t^2&=16.33\\t&=4.04 \ s\thinspace (3\thinspace s.f.)\end{aligned}$ 

The time taken to reach the ground is $\underline{4.04 \ s}$ .

To calculate the horizontal range, use $s=vt$ :

$s=20\times 4.04\\ s=80.8 \ (3\thinspace s.f.)$ 

The particle has a horizontal range of $\underline{80.8\ m}$ .