Projectile Motion

A projectile is anything that is thrown into a free trajectory.  When it's being thrown, a force is applied.  Once that force stops, the object is influenced by the pull of gravity.  Therefore we have to consider its motion in two dimensions:

1. The horizontal motion;

2. The vertical motion.

It is important to take into account the direction that the projectile is taking. 

Students get very anxious about this topic, but it's really quite simple as long as you keep your wits about you.  At this level the key point to realise is that we ignore air resistance, so:

1. The horizontal velocity does not change.

2. The vertical velocity changes in accordance with the equations of motion that use an acceleration.  On the Earth that value is quoted as 9.8 m s^-2, although 10 m s^-2 is a good enough approximation.  In reality, the acceleration is always going to be 9.8 m s^-2, as athletics on other planets or the moon are unlikely.  But that's not to rule out some daft examiner setting such a question!

Vertical Motion only

An example of this is throwing a ball into the air.  As soon as the ball is released, it is travelling vertically upwards at a velocity of +u metres per second.   The plus sign tells us it's going upwards. 

It immediately starts to slow down at an acceleration of - 9.8 m s^-2.  The minus sign tell us that the acceleration is downwards.  It is vital to get the sign right when using the equations of motion.

To explain how this works, let's do a worked example:

We can represent the vertical motion graphically.   The horizontal motion is zero.

A displacement time graph of this motion looks like this:

The velocity time graph looks like this:

Firing a Projectile Horizontally

The apparatus in this picture fires a projectile horizontally at a constant velocity from left to right.  At the same time it drops a second identical ball vertically.  Pictures are taken every second.

We see that the ball that is flying horizontally is also accelerating downward at exactly the same acceleration as the ball that is dropped vertically.

There are two components to the ball with horizontal motion:

1. Constant horizontal velocity given by

   

2. Vertical velocity given by an appropriate equation of motion.

Each component needs to be worked out separately.  The resulting motion is given by combining the two velocities to give a resultant velocity.  This is shown below:

 So if we call the horizontal velocity v_H and the vertical velocity v_V, we can work out the resultant v_R as:

Throwing a Projectile Upwards and Forwards

The physics principles are similar to the last case we looked at, except that the motion is in two phases:

1. Going up

2. Coming down.

Again the horizontal velocity remains constant because we ignore air resistance.  The vertical velocity is treated using an appropriate equation of motion.  The trajectory is shown below.  It is a parabola.

If we know the angle at which the projectile is thrown, and its velocity, we can resolve its motion into vertical and horizontal components.

So we can say that:

• Horizontal velocity, v_H is given by:

• Vertical velocity, v_V is given by:

If we want to know the range of the projectile, we need to know the time that is taken for the projectile to reach its maximum height.  In other words this  is the point where the vertical velocity is zero.  We use:

Now the velocity at the end v = 0 and initial velocity u = v_V = v sin q.  So we can write:

We can rearrange this to give:

This then gives us:

Note that the acceleration is negative (towards the ground).  Therefore the time is positive.  Wouldn't negative time be wonderful?  We could go back in time and do those things we ought to have done... 

However we must remember to double the time, as the projectile will take the same time to fall back to the ground.  So the total time in the air is now:

If we want to go the whole hog, we can now write an expression for the range:

Therefore:

We can tidy this up to write:

[Boom! Boom!]

We can mathematically model on a spreadsheet to show that the maximum range occurs when the angle of throw is 45 degrees.

This model depends on the projectile being thrown from ground level.   However real athletes would throw from a height of say 1.5 metres above the ground.  If you want a challenge, you could derive an expression by:

1.  working out the vertical displacement to give you the time (use v² = u² + 2as);

2. then working out the time for the first half of the trajectory (use s = ut + 1/2 at²);

3. then use s = ut + 1/2 at² to work out the time for the second part of the trajectory.  Remember that s in this case is the height you worked out in step 2 + 1.5 m (or whatever it is).

Definitely one for the spreadsheet!