Projectile Motion




The properties of the motion described in Section 7.2 confirm that a vector approach can be used in the analysis of projectile motion. We have already considered examples of velocity as a vector - see for example the relative velocity problems of Chapter 1. Here we will need to use vector concepts of displacement, velocity and acceleration. As with any vector quantity in two dimensions, we can resolve the motions so that they can be analysed independently in any two mutually perpendicular directions. This independence was illustrated in the coin experiment and will be used in the examples that follow.




If you have attempted the experiment in Section 7.2, then it is unlikely that your results will have been entirely conclusive. Such experiments highlight the shortcomings of idealised mathematical models. The term ”„ideal conditions”¦ will often be used. These conditions embody the following assumptions:

(a) The particle model. Projectiles are particles and, as a result, are subject to no resistance forces which depend upon their size. All resulting motion will be translational.

For real problems, modelling a body as a particle is a major assumption. A long-jumper's jump may be considered as the path of a projectile, but are we justified in using the particle model in such a case? Any body of finite size will rotate, and that rotational motion may appear to have an effect on the projectile path.

In addition, air resistance will always act as the jumper has size, and those forces of resistance usually depend upon the velocity and size of the body. The effects of resistance can be large, but will be minimised if the time of flight is kept short, the velocity is not large and the body's dimensions are small. Analysing these assumptions carefully can supply important feedback to the projectile problem.

(b) The acceleration due to gravity is a constant. This is a reasonable assumption. Clearly, launching satellites from the Earth's surface cannot be modelled as a projectile.

(c) The motion will be confined to two dimensions. This is not always the case in real example of projectile motion, as any golfer or footballer will know. There can be considerable sideways movement (swerve) created by what is known as the Magnus effect (the description of which is outside the scope of this book, but see for example texts on fluid mechanics, or S. Townend, The Mathematics of Sport, Ellis Horwood, 1987).

(d) The space in which the projectile travels is a vacuum. The inclusion of this assumption removes many of the problems of resistance already mentioned. The effects of wind and air currents would not be experienced in a vacuum, which is a prospect that would be relished by many sports men and women.

In real problems, these assumptions may not be explicitly stated. It is always advisable to list any assumptions that you feel you are making in modelling a real problem and consider the significance of each in your final solution. In this way, you will become aware of which assumptions are justifiable and which are not.

Example 7. 4

A woman is standing on a horizontally moving airport walkway which moves at a uniform speed of 2 ms-1. She notices that the walkway passes under a fixed barrier some distance ahead of her and decides she will jump the barrier when at some strategic position. She remembers that she can jump, from a standing position, a vertical height of only 1.25m, and realises that this is exactly the height of the barrier. If she jumps vertically in order to just clear the barrier, by considering her motion to be ”„ideal”¦ as described, determine:

(a) the vertical velocity with which she leaves the ground;

(b) the time she will be airborne;

(c) the distance from her take-off position, on the walkway, that she touches down;

(d) the distance her touch-down position has moved horizontally during the motion.

(We will assume that the acceleration due to gravity is 10 ms-2.)

Solution Figure 7.2 defines the relevant quantities.

(a)Consider the vertical motion: At her greatest height, the woman will have zero vertical velocity; thus

0 = v2 ”V 2 X 10 X 1.25

where v is her initial vertical velocity. It follow that v = 5ms-1

(b)Consider vertical motion: The woman will be airborne for t seconds until she is again at zero vertical height; thus:

0 = 5t ”V 0.5 X 10 X t2

The two solutions here give t = 0 at take-off and t = 1 at landing. The time airborne is thus 1s.

(c)Consider horizontal motion: Both the woman and the walkway have the same horizontal velocity; she will, as a result, touch down at the same place on the belt that she taken off.

(d)Consider horizontal motion: Horizontal distance = horizontal velocity X time airborne. The distance moved horizontally is thus 2 X 1 = 2m.