Projectile Motion Graphs.

NB: The vertical projectile motion being discussed here is in one dimension.

In this post, we will be explaining the projectile motion graphs. There are three kinds of graphs that you will be required to draw in this section:

1. Velocity vs. time
2. Position vs. time
3. Acceleration vs. time graphs

The motion we will be looking at is when the projectile is not bouncing when it lands on the ground/surface.

To be able to understand the graphs that we will show you below, you need to have prior knowledge of functions in grade 11 maths. We relate these functions, especially the linear function and the parabola with the equations of motion:

\begin{alignat} {1} v_f &= v_i +a\Delta{t} \tag{1} \\ v_f^{2} &= v_i^{2} + {2}{a}\Delta{t} \tag{2} \\ \Delta{y} &= v_i\Delta{t} + \frac{1}{2}a\Delta{t^2} \tag{3} \\ \Delta{y} &= (\frac{v_i+v_f}{2})\Delta{t} \tag{4} \\ \end{alignat}

As you continue reading, we will show you how these equations are related to their maths functions we mentioned above.

Velocity vs. time graphs.

To plot the graph of velocity vs time, we use the equation:

\begin{gather} v_f=v_i + \Delta{at}\\ y=c+mx \end{gather}

As you can be see from the equations above, the equation of motion relating velocity to time represents some straight line function. The initial velocity, \(v_i\) represents the y-intercept \(c\) and the gravitational acceleration, \(a\), represent the gradient \(m\).

Velocity time graph when a ball is dropped (upwards is positive)

From the graph above, it is clear that it was dropped because \(v_i\) (y-intercept) is zero. The slope of the line is negative and this is true because upward movement is assumed is positive and therefore gravitational acceleration which is a downward motion is negative.

Let us now see how the graph will look like when a projectile is thrown upwards and returns to point of release.

Velocity time graph when a ball is dropped (upwards is positive) Velocity vs. time graph when a projectile is thrown upwards (downwards is positive

The graphs are a representation of a projectile thrown upwards and returns to the thrower's hand. When the projectile returns to the thrower, the arms of the graph will be equal. By this we mean, top part will equal the bottom if your graph is drawn to scale.

What happens if the projectile passes the thrower's hand and hits the ground?

In this case, the arm that corresponds to the ground will be longer. This is a situation that will help you see if a projectile returned to the thrower's hand or passed to hit the ground. Pay attention to these details.

Area under the graph.

You will be asked to calculate the area under a curve/graph. Sometimes they even ask you what property does the area represent. Let's refer to the graph below.

Tiisetso throws the ball upwards at 20ms-1. This is the graph that represents the motion of the ball.

Now let us assume the examiner asks you what is the area under the graph at t = 1s and what quantity does this represent.

To answer the question, we first ask ourselves, if we calculate area it means the multiplication of two opposing sides. This in our case means multiplying velocity with time. The quantity would be displacement.

Calculating the area, we see that we must calculate the areas of two shapes, i.e., a triangle and a rectangle. Look at the area of the graph at t = 1s. From \(v = 10 - 20ms^{-1}\) and \(t = 0 - 1 s\), the shape is a triangle. For \(v = 0 - 10ms^{-1}\) and \(t = 0 - 1s\), the shape is a rectangle. Therefore, the area is is the sum of these two shapes:

\begin{gather} A=\frac{1}{2}{b}{h}+lb\\ =\frac{1}{2}(\,1)\,(\,10)\,+(\,1)\,(\,10)\,\\ =15m \end{gather}

Position vs. time

Now we have to look at the equations of motion that can relate position with time. The following equation is the one that can relate position with time.

\begin{gather} \Delta{y}= {v_i}{\Delta{t}}+{\frac{1}{2}}{a}{\Delta{t^2}}\\ y=bx+ax^2 \end{gather}

When you look at the equations above, you realise that this is a parabola. However, the y-intercept is zero. The meaning of this is that when you plot a graph, it is assumed that the reference point is the point of release of the projectile. Look at the graph below to understand.

Position vs. time when an object is dropped(upwards is positive).

The graph above starts from \(y=0\). This is the point of release of the projectile. Remember that the projectile is dropped, meaning its initial velocity is zero. The upwards motion is chosen as positive, and this makes gravitational acceleration negative. Hence the concave down shape (frowning face). The mathematics of this shape is:

\begin{gather} \Delta{y}= {v_i}{\Delta{t}}+{\frac{1}{2}}{a}{\Delta{t^2}}\\ \Delta{y}= 0 \times {\Delta{t}}+{\frac{1}{2}}{(\,-9.8)\,}{\Delta{t^2}}\\ \Delta{y}= {-4.9}{\Delta{t^2}} \end{gather}

The mathematics above is just to help you understand how these graphs are drawn. They are not for exam purposes.

The other position-time graphs are as follows:

Position vs. time graph (reference point is the point of release) when the projectile is thrown upwards and passes the initial position to hit the ground. (upwards is positive).

The arm of the curve on the side of the opposite motion (in this case, downwards motion) will become longer if the projectile passes the point of release, to hit the ground. When the point of release is chosen as the reference point, the point becomes the origin point of the graph.

Position vs. time graph (reference point is the ground) when the projectile is thrown upwards and passes the initial position to hit the ground (upwards is positive).

When the ground is taken as a reference point, the graph is shifted up by that height.

Acceleration vs. time

When objects are in free fall, the only force that is affecting them is the force of gravity and therefore, the gravitational acceleration will remain constant. We therefore expect the graph to be some constant line. The graphs are given below.

Acceleration vs. time (upwards is positive). Acceleration vs. time (downwards is positive).