# Full-flight Projectile Motion: Initial Velocity and Launch Angle

This topic is part of the HSC Physics syllabus under the section Projectile Motion.

### HSC Physics Syllabus

• Apply the modelling of projectile motion to quantitatively derive the relationships between the following variables:

– initial velocity
– launch angle
• Solve problems, create models and make quantitative predictions by applying the equations of motion relationships for uniformly accelerated and constant rectilinear motion

### Initial Velocity and Launch Angle

• All objects at the beginning of their projectile motion must possess a non-zero initial velocity.
• The initial velocity can always analysed as and resolved into two components: horizontal and vertical velocities. This is done by constructing a right-angled triangle from vectors.

Conventions for calculations
• Initial velocity is typically written as u
• Initial vertical velocity is typically written as u_y (subscript y is used to represent the vertical rectilinear motion)
• Initial horizontal velocity is typically written as u_x (subscript x is used to represent the horizontal rectilinear motion)
• The relationship between initial velocity, initial horizontal and vertical velocity can always be represented by the right-angled triangle where \theta (as shown in the diagram) is the launch angle at which the projectile leaves the horizontal plane (usually the ground).

Using trigonometry, initial horizontal and initial vertical velocities can be expressed in terms of the initial velocity.

Initial horizontal velocity:

u_x/u=costheta

Initial vertical velocity:

u_y/u=sintheta

• The relationship between initial vertical and horizontal velocity is described by:

(u_y)/(u_x)=tantheta

• Since the triangle is right-angled, the three vectors’ relationship can also be summarised by Pythagoras’s theorem.

u^2=(u_x)^2+(u_y)^2

u=sqrt((u_x)^2+(u_y)^2)

The initial velocity can be negative because the initial direction of a projectile can also be downwards as shown below. Situations in which this type of initial velocity occurs will be explored and clarified in practice questions later.

### Example 1

A projectile is launched at 60 m s-1 at an elevation of 30º. Find its initial horizontal and vertical velocities.

Solution:

Construct a right-angled triangle from vectors:

Initial horizontal velocity:

u_x=60cos30^o

Initial vertical velocity:

u_y=60sin30^o

• Initial vertical component of velocity changes throughout projectile motion. Its magnitude decreases when a object travels upwards and increases when it travels downwards.
• Initial horizontal component of velocity remains constant and does not change (assuming absence of air resistance).

### Changes in Vertical Component of Velocity in Projectile Motion

Motion in the vertical axis can be modelled using rectilinear equations. In contrast, motion in the horizontal axis does not require these equations because horizontal acceleration is zero.

s=u_yt+1/2a_yt^2

v_y=u_y+a_yt

v_y^2=u_y^2+2a_ys

where:

• s represents the vertical displacement
• a_y represents the acceleration in the vertical axis (gravity)
• u_y and v_y are the initial vertical velocity and velocity after time t of the object during its projectile motion.

### Example 2

The initial vertical velocity of an object during projectile motion is 15 m s-1. The launch angle is 30º.

(a) Calculate the horizontal component of the object's initial velocity.

(b) Calculate the vertical component of velocity and the instantaneous velocity 2 seconds after the object’s launch.

Solution to part (a):

u_y/u_x=tantheta

u_x=u_y/tantheta

u_x=15/tan30^o

u_x=25.98  ms^-1

Solution to part (b):

The vertical velocity can be modelled by the following equation:

v_y=u_y+a_yt

Since, acceleration due to gravity is acting against the direction of the object, a negative sign must be placed in front of 9.8.

v_y=15+(-9.8)(2)

v_y=-4.6  ms^-1

The instantaneous velocity can be determined using Pythagoras’s theorem:

v=sqrt(v_x^2+v_y^2)

v=sqrt((25.98)^2+(-4.6)^2)

v=26.4  ms^-1

Since velocity is a vector quantity, a direction is needed. The direction is typically indicated by the angle relative to the horizontal.

Using trigonometry:

v_y/v=sintheta

4.6/26.4=sintheta

theta=sin^-1(4.6/26.4)

theta=10.0^o

Therefore, the instantaneous velocity of the object 2 seconds after launch is 26.4 ms-1 10.0° relative to the horizontal axis.

Note that the construction of a right-angled triangle using velocity vectors can be done at any point during an object’s projectile motion. The horizontal velocity (horizontal vector) remains constant and does not change in length. The angle, vertical component of velocity (vertical vector) and the instantaneous velocity (hypotenuse) change with time.

Previous section: Introduction to Projectile Motion

Next section: Full-flight Projectile Motion - Maximum Height, Time of Flight and Range