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Projection at any angle

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Summary

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 $U \cos \alpha$ .

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)

Learn with Basics

Length:

Unit 1

Equations of motion

Unit 2

Projectile motion

Jump Ahead

Optional

Unit 3

Projection at any angle

Final Test

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FAQs - Frequently Asked Questions

How does the horizontal component of velocity change when a particle is projected at an angle?

The horizontal component of velocity does not change, it remains constant, assuming there is not air resistance.

When a particle is projected at an angle, what happens at the greatest height?

At the greatest height, the vertical component of the velocity is zero.

Can a particle be projected at any angle?

Yes, a particle can be projected horizontally, vertically or at any other angle.

Projection at any angle

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Summary

Exercises

Select Lesson

Exam Board

Further kinematics

Application of forces

Projectiles

Forces and friction

Moments

Forces and motion

Variable acceleration

Constant acceleration

Modelling in mechanics

Normal distribution

Conditional probability

Correlation and hypothesis testing

Hypothesis testing I

Binomial distribution

Probability

Representation of data

Measures of location and spread

Data collection

Vectors II

Numerical methods

Integration II

Differentiation II

Trigonometry and modelling

Trigonometric functions

Radians

Sequences and series

Binomial expansion II

Parametric equations

Functions

Algebraic fractions

Proof II

Vectors I

Integration I

Differentiation I

Exponentials and logarithms

Trigonometric equations

Trigonometry

Binomial expansion I

Circles

Straight line graphs

Transformations

Sketching graphs

Polynomials

Equations and inequalities

Quadratics

Indices and surds

Proof I

Explainer Video

Summary

Projection at any angle

In a nutshell

Projection at any angle

Horizontal motion

Vertical motion

Equations

description

equation

Variable definitions

​​quantity name

symbol

unit name

unit

Example

Learn with Basics

Unit 1

Equations of motion

Unit 2

Projectile motion

Jump Ahead

Unit 3

Projection at any angle

Final Test

FAQs - Frequently Asked Questions

How does the horizontal component of velocity change when a particle is projected at an angle?

quantity name

quantity name