Analysing Projectile Motion - Learn

Analysing Projectile Motion – Learn

Objects that are close to the Earth’s surface accelerate toward the surface of the Earth as a result of their weight force. The weight force is due to the Earth’s gravitational field which is considered a constant for objects that are near the Earth’s surface ( $g=9.8\: { m }/{ s^{ 2 } }$ ).

As g is constant for all objects, all objects must accelerate toward the Earth at the same rate. This is not apparent in real life (consider dropping a feather and a rock at the same time – a rock will hit the ground first). This is because there is a net force acting on any object falling toward the Earth. This net force is a result of the weight force of the object acting down and a resistive force acting upward. The resistant force that opposes the motion of a falling object is known as air resistance or drag ( ${ F }_{ drag }$ ). The drag force increases as the objects speed increases.

The drag that an object experiences as it accelerates toward Earth is proportional to:

The area of the object perpendicular to the direction of motion
The density of the air through which the object moves
The objects speed

The effect of drag on a projectile can be observed with the simulation at the bottom of the page.

The net force acting on an object falling toward the Earth under the influence of gravity is:

${ F }_{ net }\: =\:W+{ F }_{ drag }$

Terminal velocity

An object that accelerates toward Earth experiences a constant weight force, W. The drag force ( ${ F }_{ drag }$ ) that the object experiences, increases to a point where it is equal but opposite in direction to the weight force. At this point the net force ( ${ F }_{ net }$ ) acting on the object is zero and it will no longer accelerate and its velocity remains constant. This velocity is called the terminal velocity.

Analysis of Projectile Motion

Characteristics of projectile motion:

An object is given an initial velocity
The object continues to move due to its own inertia and its weight force
The projectile follows a parabolic path (ignoring air resistance)

When we analyse projectile motion problems, we make two assumptions:

All objects experience a constant vertical acceleration due to gravity ( $g=9.8\: { m }/{ s^{ 2 } }$ )
All objects experience zero air resistance

Modelling Projectile Motion

The following assumptions are made for all projectiles analysed in this course:

For projectiles launched horizontally the initial vertical velocity is 0
For projectiles launched vertically the initial and final horizontal velocity is 0
The horizontal component of the projectiles velocity is constant
The vertical component of the projectiles velocity constantly changes

The diagram below illustrates how the horizontal and vertical velocity vectors vary over time. Note that the horizontal vector is constant and the vertical vector is zero at the top of the trajectory and it changes direction and magnitude:

When modelling projectile motion problems the horizontal and vertical vector components are analysed separately. The diagram below illustrates how the initial velocity, u, is resolved into its horizontal and vertical components:

This is usually simplified to the diagram below for analysis:

Example 1:

A 75kg skydiver jumps from a plane and accelerates toward the Earth.

a) What is the acceleration of the skydiver if we assume a constant drag force of 60N?

b) What is the skydivers vertical speed 5 seconds after jumping out of the plane?

Answers:

a) ${ F }_{ net }\: =\:W+{ F }_{ drag }$

${ F }_{ net }\: =\: (75\times 9.8)+(-60)$

${ F }_{ net }\: =\: 675N$ (net force experienced by the sky diver)

$ma\: =\: 675N$

$75a\: =\: 675N$

$a\: =\: \cfrac { 675 }{ 75 }$

$a\: =\: 9\:{ m }/{ { s }^{ 2 } }$

b) Using: $v\:=\:u + at$

Where:

$v\:=\: ?$

$u\:=\:0\:{ m }/{ s }$

$a$ = $9\:{ m }/{ { s }^{ 2 } }$

$t\:= \:5 s$

$v\:=\:0+9\times 5$

$v\:=\:45\:{ m }/{ s }$

Example 2:

A projectile is launched with an initial velocity of $25\:{ m }/{ s }$ at an angle of ${ 30 }^{ \circ }$ to the horizontal. Calculate the horizontal and vertical components of the projectiles velocity:

Horizontal:

${ u }_{ x }=ucos\theta$

${ u }_{ x }=25cos30$

${ u }_{ x }=21.65\:{ m }/{ s }$

Vertical:

${ u }_{ y }\:=\:usin\theta$

${ u }_{ y }\:=\:25sin30$

${ u }_{ y }\:=\:12.5\:{ m }/{ s }$