Students often find it difficult to calculate the trajectories of projectiles. With the help of Elias Kalogirou’s model, they can be easily visualised. In addition, Ian Francis suggests further uses for the model in the classroom.
When you throw or hit a ball, shoot a bullet from a gun or drop a stone from a bridge, the ‘flying’ objects all have one thing in common: in the physical sense, they are projectiles. This term is used for any object that is given an initial velocity and subsequently follows a distinct trajectory: a path determined by a combination of gravity and air resistance. I have devised a model to help visualise these trajectories in the classroom and allows students to investigate horizontal and vertical components of projectile motion. Note that air resistance is not included in the model.
The model is now complete. As a guide, it took me about two hours to build.
To understand how we calculate the lengths of the strings, we need to understand what the model represents. Imagine that at time zero, you fired a bullet at steady speed (no horizontal acceleration) from the pivot point (connecting the ruler and the clamp stand), in the direction that the ruler is pointing in.
The model demonstrates two aspects of a trajectory. Firstly, the direction in which the ruler is pointing gives the direction in which the bullet would continue flying if there were no gravity.
Secondly, the strings represent the effect of gravity (g). If you let the bullet drop from the pivot point at time zero (vertical fall without initial velocity), the length of string of the first bead would give the distance the bullet would have fallen after time t, the string of the second bead would give the distance the bullet would have fallen after 2t, and so on (see Table 1).
Gravity has the same effect on a projectile with an initial velocity greater than zero, so when you shoot the bullet rather than letting it drop, it would still fall the same distance by time t.
For an initial velocity greater than zero, the length of the first string again gives the distance the bullet would have fallen after time t, the second string gives the distance the bullet would have fallen after 2t, and so on. The beads, therefore, represent the parabolic trajectory of a projectile, with the angle of the ruler to the stand representing the starting angle of the projectile.
The positions of the 20 beads, hanging 5 cm apart, give you the positions of the bullet at 20 equidistant consecutive time points – the first bead at time t, the second bead (5 cm further along the ruler) at 2t, and so on, up to the last bead, at 20t.
The model represents trajectories at constant horizontal velocity (including zero, if you place the ruler in a vertical position, parallel to the stand) and constant vertical acceleration. Once you have chosen your value for t, cut the strings and built the model, it will be a model for trajectories with this specific value of t, i.e. also for a specific velocity and acceleration (gravity) – at different angles, depending on how you position the ruler, and disregarding air resistance. How closely the model reflects reality could be an interesting point for discussion with your students.
The length of the shortest string, at time t, is calculated as:
a = ½ g t2.
To calculate its length (a), choose the maximum length of string for bead number 20, which is 100 cm from the pivot point, and corresponds to 400a (see Table 1). Our longest string (400a) was 145 cm, so a = 0.3652. Then you can calculate the lengths of string you need to cut for the 20 different beads using the ‘String length’ column in Table 1 (and do not forget the 5 cm extra when cutting, see step 4 above).
|Time||Distance along the ruler (cm)||Distance fallen||String length||String length if a = 0.3652 (cm)|
|t||5||½ g t2||a||0.3652|
|2t||2 x 5=10||½ g (2t)2||4a||1.4608|
|3t||3 x 5=15||½ g (3t)2||9a||3.2868|
|4t||4 x 5=20||½ g (4t)2||16a||5.8432|
|20t||20 x 5=100||½ g (20t)2||400a||145|
The actual construction of the model is already a valid act of learning in itself. In addition, the finished product can be used for further work. It can be used to study either horizontal or vertical trajectories, as well as those at any angle in between. Below are some suggestions – there will be plenty of others.
In this experiment, students learn that the horizontal and vertical components of a trajectory are independent of each other, with the horizontal velocity remaining constant during the ‘flight’.
Students may be familiar with the average velocity formula, but less so with the idea of dividing up a motion into small time intervals. Therefore, it could be worth having them calculate the average horizontal velocity = total horizontal displacement / total time. This should, of course, equal the velocities worked out from adjacent bead positions.
In this experiment, students learn that the horizontal velocity will still be constant for a trajectory with initial vertical velocity (i.e. at an angle away from the horizontal), but it will be smaller than that for a trajectory with no initial vertical velocity (as in Experiment 1).
Again, if the model has been built accurately, the students should find that the horizontal distance travelled in equal intervals of time is constant, but it will of course be smaller than that for a trajectory with no initial vertical velocity (Experiment 1) – the strings of adjacent beads will be closer to one another.
In this experiment, the students study the vertical distances travelled in equal intervals of time for a trajectory with no initial vertical velocity (such as in Experiment 1). The experiment is best suited to students who were not involved in building the model, although it can be useful reinforcement for those students, too.
A change in the vertical velocity is an acceleration (a), and from the equation F = m a we know that a resultant force is needed to produce such an acceleration – in this case the force of gravity acting on the object. As this force is constant, from equations of uniformly accelerated motion, we get
vertical distance travelled (s) = (u t) + (½ a t2)
where u = initial velocity.
As the initial vertical velocity (u) is zero, (u t) can be ignored, and as ½ a is a constant, the relationship tells us that vertical distance travelled is proportional to time squared. It may be worth pointing out that acceleration (a) and gravity (g) are interchangeable in this context, both representing the acceleration of freefall.
Instead of using the value for t built into the model, students could use the value calculated in Experiment 1. If this does not correspond to the value of the model, the graph will still be the expected straight line, showing the same correlation, but only above the second point in the graph.
This simple experiment serves to reinforce the fact that the vertical and horizontal components of a velocity are independent of each other.
These are further questions you can ask the students to investigate: