## Projectiles

A projectile is a particle that is launched and then moves by the action of its own weight. When dealing with projectile motion we take air resistance to be negligible. This therefore means the only force acting on a launched projectile is the force of gravity.

## Components of a projectile

When a projectile is launched at any angle other than 90° to the horizontal its motion can be resolved into separate vertical and horizontal components.

## Initial velocity

The initial velocity is the velocity with which the projectile is launched. By convention we assign it the symbol (u). Since the initial velocity of projection is at an angle to the horizontal, we resolve it into a horizontal component (ux) and a vertical component (uy) as follows:

Velocity is a vector. According to the laws of vectors, parallel vectors of the same magnitude are equal. So in order to be able to calculate the magnitudes of the vertical and horizontal components of the initial velocity, we shift the vertical component to the right in such a way as to complete the vector triangle. This is illustrated below:

The vector triangle created is a right angled triangle in which:

• the initial velocity (u) is the hypotenuse
• the vertical component (uy) is the opposite
• the horizontal component (ux) is the adjacent

Using trigonometrical ratios, we can calculate the components as follows:

substituting in uy and u we get:

making uy the subject of formula to get the formula for uy:

We can also do the same using the cosine ratio:

Therefore:

• initial vertical velocity (uy) = u sin θ
• initial horizontal velocity (ux) = u cos θ

## Velocity after time (t)

For a projectile, we can find the velocity after time (t) using the basic equations of motion. Projectile motion is parabolic motion that can be resolved into vertical linear motion and horizontal linear motion. So when calculating final velocity we consider the two components of motion separately.

• initial velocity = u sin θ as we proved earlier on.
• acceleration = -g. Whereby g is the acceleration due to gravity. We use the negative value of g because the projectile is launched opposite to the direction of action of the force of gravity.

Now lets take this equation of motion:

and substitute the vertical components of the projectile to get:

vx is the vertical velocity of the projectile after time (t)

Now lets move on to the horizontal components of the motion.

• initial velocity = u cos θ
• acceleration = 0 m/s2. This is because, horizontally there is theoretically no acceleration or deceleration because we assume air resistance to be negligible.

vy is the horizontal velocity of the projectile and as you can see from the formula, it is constant and equal to the initial horizontal velocity.

The next diagram is a visual representation of what we have learnt so far.

The diagram above shows a projectile that was launched at O. Point A represents a random point after time t along the path of the projectile.

Now that we know how to find the components of the velocity after time (t), velocity (v) can be found by finding the resultant of the components.

### Worked example

A ball is thrown with a velocity of 30 m/s at an angle of 30° to the horizontal. Find its velocity after 3 seconds.

### Solution

We know that:

ux = 30 cos 30°

uy = 30 sin 30°

From there we can calculate the vertical velocity after 3 seconds as follows:

vy = 30 sin 30° – 10(3)

= 15 – 30

= -15 m/s

horizontal velocity after 3 seconds:

vx = 30 cos 30°

= 26 m/s

After we find the horizontal and vertical velocities, we then find the resultant velocity of the two. The diagram below will help you understand what I mean.

After 3 seconds, the projectile is moving like this:

We then use Pythagoras theorem to find the resultant velocity:

Since velocity is a vector, we give both magnitude and direction:

The vertical velocity is negative but when finding the angle we use its absolute value which is positive:

Ans: the velocity after 3 seconds = 30m/s at -30° to the horizontal.

## Displacement

Knowing the vertical and horizontal components of the velocity of a projectile we can find its displacement from the point of projection after time (t).

As usual, we derive our formulae from the equations of linear motion.

First we are going to start by deriving the equation for vertical displacement using the formula:

Then we find the equation for horizontal displacement using the same equation of motion.