Projection at any angle

In a nutshell

Now that you have learnt how to resolve the initial velocity into horizontal and vertical components, it is possible to tackle projectile motion problems where the particle is projected at any angle. When considering the motion of a particle projected at an angle, the vertical and horizontal components are independent of each other, and therefore the calculations for vertical and horizontal movement can be considered separately.

Projection at any angle

When a particle is projected at an initial velocity $U$ at an angle $\alpha$ above the horizontal, resolve the initial velocity into components. The horizontal and vertical motion can then be considered separately.

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. The horizontal component of the initial velocity is $U \cos \alpha$ .

Maths; Projectiles; KS5 Year 13; Projection at any angle

Vertical motion

Acceleration due to gravity is $9.8 \ m.s^{-2}$ , and so the formulae for constant acceleration can be used in the vertical direction. The vertical component of the initial velocity is $U \sin \alpha$ . When a projectile reaches its greatest height, the vertical component of the velocity is $0$ .

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

Example

A particle is projected from $3\ m$  above the horizontal ground with a velocity of $(4i+6j)\ m.s^{-1}$ . Take the upwards and right each to be positive. Find:

i) The greatest height above the ground.

ii) The speed with which the particle hits the ground.

iii) The angle of the particle as it hits the ground.

i) To calculate the greatest height of the particle, consider the vertical motion of the particle:

$s=\space? \qquad u = 6 \qquad v=0 \qquad a = -9.8 \qquad t=\space?$ 

$\begin{aligned}v^2&=u^2+2as\\0^2&=6^2+2\times (-9.8) \times s\\s&= 1.84\ m\end{aligned}$

Since the particle was launched from $3\ m$ above the ground, the greatest height reached is:

$\underline{3+1.84=4.84\ m} \ (3\thinspace s.f.)$

ii) The horizontal component of the velocity is always $4\ m.s^{-1}$ .

The vertical component can be calculated:

$\begin{aligned}v^2&=u^2+2as\\v^2&=6^2+2\times(-9.8)\times (-3)\\v^2&=94.8\\v&=\pm9.7365\ m.s^{-1} \end{aligned}$ 

The speed is therefore:

$v=\sqrt{4^2+9.7365^2}=10.526\ m.s^{-1}$

The particle hits the ground at $\underline{10.5\ m.s^{-1} \ (3 \thinspace s.f.)}$ .

iii) The angle of this collision is given by:

$\tan{\theta}=\cfrac{9.7365}{4}\implies\theta=\underline{68\degree}$ (to the nearest degree)