Topic 3 - Projectiles

A projectile is any object that is thrown by any means.  There are three situations we will look at:

• Throwing an object vertically into the air;

• Throwing an object horizontally from a height;

• Throwing an object at an angle.

  The key concepts you must understand:

• The horizontal movement is totally independent of the vertical movement.  That means that they do NOT affect each other.

• The two movements are vector quantities, so they have a direction.

• The velocities are at 90^o to each other.

• There is a resultant velocity from the two independent velocities.

• We can analyse the vertical movement using the equations of motion.

Throwing An Object Vertically Into The Air

Consider a basketball player throwing a ball in the air.  What goes up must come down.

The ball has a downward force acting on it because of gravity.  Therefore it will slow down at a rate of 10 m/s².  So we can say that the acceleration is -10 m/s².  When we tackle problems like this, we use the equations of motion. Click HERE if you need to review the equations of motion. We have to make sure that we get the signs right.  We will make upwards positive and downwards negative.

We can represent these motions graphically.  It is important that you understand these graphs.

A displacement time graph looks like this:

The graph is a parabola because the ball is accelerating downwards.  When it reaches the top, its velocity is 0, but its acceleration is still (-)10 m/s². 

The velocity time graph looks like this:

This graph shows how the velocity not only changes, but its sign changes as well.  This tells us that the direction changes as well.  This stands to reason; if going up is positive, going down must be negative.  Note that the gradient is constant, i.e. the acceleration is constant.

The acceleration time graph looks like this.

It shows us that the acceleration is constant at -10 m/s².  The minus sign tells us that the acceleration is towards the ground.

Throwing an object horizontally from a height

If we throw an object horizontally, there are two important things to consider:

• The horizontal velocity remains constant (ignoring air resistance)

• The vertical velocity increases at a rate of (-)10 m/s².

If we throw one object and drop a second object at the same time, we see this:

The second object, thrown horizontally, will hit the ground at the same time as the object that is simply dropped.

Although the drawing is not to scale, you can see how the horizontal velocity remains constant, while there is acceleration downwards.

Look at the diagram below.  A pallet is dropped from a helicopter to the ground.  We will ignore the air resistance.

The path taken is NOT a straight line, because the velocity downwards is increasing at a constant rate of 10 m/s².  It is a parabola.  There are two components in this problem:

• The horizontal velocity  which remains constant.

• The vertical velocity which changes, as the object is accelerating towards the ground. We use an equation of motion to analyse the motion.

The key point to remember is that the horizontal and the vertical motions are independent. 

A common bear-trap is to put the horizontal velocity into the vertical equation of motion.

Throwing an object at an angle

Archery is a sport in which the participants subconsciously do calculations involving movement in two directions.  Again the vertical and horizontal movement are independent.

Let us analyse the motion from the moment an arrow is released to the moment it hits the target.  We want to find the range.  For simplicity we will assume that the target is at the same height as the release point.  We will also ignore air resistance.  We will not worry about the signs.

1.  On release, the arrow leaves at a velocity v m/s and angle q.  The horizontal velocity is v cos q m/s.  The horizontal velocity remains constant at v cos q metres per second.

2.  The vertical velocity is v sin q metres per second initially.

3. To work out the time we need to use an equation of motion that has initial velocity, acceleration, and time.

    v = u + at will fit the bill.

    0 = v sin q + at

    v sin q  = at

Therefore t = v sin q

                        a

4. To get the range we need to multiply the horizontal velocity by the time taken in the air. Therefore:

range = v cos q  × 2t

It is 2t because it takes t seconds for the arrow to go up to its maximum height and t seconds for it to come down again.

In the AS exam, they will not be over officious with signs, but make sure you explain each step.