Projectiles

Examples of projectiles

Real-life examples of projectile motions are:

• The movement of a bullet shot from a gun.

• The movement of cannonballs.

• Throwing a javelin.

• Casting a stone.

• Jumping into a swimming pool.

Types of projectile motion

There are two types of projectile motion – horizontally shot projectiles, and non-horizontally shot projectiles.

Horizontally shot projectiles

This takes place when an object is launched in a horizontal direction from a height above the ground. From this elevation, the object follows a curved path before hitting the ground. An example of a horizontally shot projectile is a bullet shot from a gun in the horizontal direction.

Non-horizontally shot projectiles

This occurs when an object is launched from the ground or from a height above the ground upwards, giving rise to a complete parabolic motion. An example of a non-horizontally shot projectile is a rocket projected vertically from the ground.

Components of a projectile

Projectile motion is best understood when split into two components: horizontal and vertical.

If a ball B is launched from the ground at a point G, the position of the ball along the curved path before it hits the ground can be given as the coordinates (x, y).

Remember that the ball moves at velocity v, and at an angle θ to the horizontal. To find the horizontal component of the velocity x and vertical component y, we apply Pythagoras' theorem. Then:

\[v_x = v \cdot \cos \theta\]

The equation \(v_x = v \cdot \cos \theta\) is regarded as the horizontal component of the velocity in the projectile motion while the equation \(v_y = v \cdot \sin \theta\) is the vertical component of the velocity.

To further determine the horizontal component of the motion, we apply SUVAT equations.

Remember:

\[S = ut +\frac{1}{2} at^2\]

So, the horizontal component of a projectile motion is:

\[x = u \cos \theta \cdot t + \frac{1}{2} \cdot 0 \cdot t^2\]

Remember that the velocity u is replaced with the horizontal component of the velocity of the object.

Thus: \(x = ut \cdot \cos \theta\) is the equation of the horizontal component of a projectile motion.

However, to determine the vertical component, you should note that the object initially moves upwards. This means that its acceleration is in the opposite direction of gravity, so:

\[a = - g = -9.8 \frac{m}{s^2}\]

Also, note that the value of u in this case is the vertical component of the velocity of the object. So, the vertical component of a projectile motion is:

\[y = u \sin \theta \cdot t + (-g) \cdot \frac{1}{2} t^2\]

Thus: \(y = ut \sin \theta -g \frac{1}{2} t^2\) is the vertical component equation of the projectile motion.

Understanding these equations along with a basic knowledge of the general equations of motion is important.

Time of flight, range, and maximum height

In projectile motion questions, you will need to calculate the range, maximum height, and time of flight. A step-by-step approach helps you understand how to solve these in this text.

What is the time of flight of a projectile?

The time of flight of a projectile is the total time taken from the point at which the object is launched to the point it hits the ground. It is denoted with the symbol T.

Recall that: \(v - u = at\) is the first equation of motion. Note that the time of flight is the time taken for the object to go up and down until it hits the floor. We have to consider the vertical component of the projectile motion.

To best understand the motion, we divide it into two levels. The first level covers when the object is launched from its starting point (the ground) to the maximum height. Then, the second level covers the motion from the maximum height to the point it drops to the launch level (the ground).

\(v_y - u_y = -g \cdot t_1\)

Remember that: \(v_y = 0, \space u_y = u \sin \theta\)

\(-u \sin \theta = -gt\)

\(t = \frac{u \sin \theta}{g}\)

Thus, time of flight T is:

\(T = \frac{ 2u \sin \theta}{g}\)

What is the range of a projectile motion?

The range of a projectile motion is known as the total horizontal distance travelled by the object from the point it is launched, to the point it hits the ground. R is used to represent the range. The range is calculated by finding the product of the velocity at the horizontal direction and the time of flight.

\(R = u_x \cdot T\)

Remember that: \(u_x = u \cos \theta\).

Do not forget the range is the horizontal distance of the projectile.

Note that:
\(T =\frac{2u \sin \theta}{g}\)

\(R = u \cos \theta \cdot \frac{2 u \sin \theta}{g}\)

In trigonometry: \(\sin (2\theta) = 2\sin\theta \cos\theta\)

Thus: \(R = \frac{u^2 \sin(2 \theta)}{g}\)

As the formula in calculating the range contains 2θ it means that the highest value of the range at 45°, which means \(\sin (2 \theta) = sin(90^\circ) = 1\).

What is the maximum height of a projectile?

The maximum height is the highest point the object reaches while moving upwards before it begins to drop. At the maximum height, the object suspends shortly before it starts a downward movement. At this level, the velocity is zero in the vertical direction.

In order to calculate the maximum height, the velocity in the vertical direction is \(u_y = u \sin \theta\).

Remember the equation of motion: \(v^2 - u ^2 = 2a \cdot S\)

Where: v = 0, S = h, a = -g.

\(-u_y^2 = -2gh\)

So:

\(h = \frac{y_y^2}{2g}\)

\(h = \frac{(u \sin \theta)^2}{2g}\)

Thus, the maximum height of a projectile is:

\(h = \frac{u^2 \sin^2 \theta}{2g}\)

How to calculate projectiles using vectors

You solve projectile questions using vectors by understanding the quantities which are vectors. In this case, only the acceleration, velocity and displacement are vectors. For vector quantities, the horizontal component i and vertical component j are always represented.

For a vector 5i, it means that the quantity is 5 units in the positive horizontal direction (to the right on the x-axis). But for a vector -3j, it means that the quantity is 3 units in the negative vertical direction (in a downwards direction on the y-axis).

The vector direction is given in place of an angular direction in degrees. You will be relying on the displacement and velocity equations:

\(S = S_0 + ut + \frac{1}{2} at^2\)

\(v = u + at\)

Where S = displacement,

S₀ = initial displacement

a = acceleration

t = time,

v = final velocity and

u = initial velocity.