Describing Projectile Motion

The following example illustrates how to mathematically describe projectile motion.

Example: Describing Projectile Motion

Question

A cricketer hits a cricket ball from the ground and the following graph of velocity vs time was drawn. Upwards was taken as positive. Study the graph and follow the instructions below:

Describe the motion of the ball according to the graph.
Draw a sketch graph of the corresponding position-time graph. Label the axes.
Draw a sketch graph of the corresponding acceleration-time graph. Label the axes.

Step 1: Describe the motion of the ball

We need to study the velocity-time graph to answer this question. We will break the motion of the ball up into two time zones: $t=\text{0}\text{ s}$ to $t=\text{2}\text{ s}$ and $t=\text{2}\text{ s}$ to $t=\text{4}\text{ s}$.

From $t=\text{0}\text{ s}$ to $t=\text{2}\text{ s}$ the following happens:

The ball starts to move at an initial velocity of $\text{19.6}$ $\text{m·s $^{-1}$}$ and decreases its velocity to $\text{0}$ $\text{m·s $^{-1}$}$ at $t=\text{2}\text{ s}$. At $t=\text{2}\text{ s}$ the velocity of the ball is $\text{0}$ $\text{m·s $^{-1}$}$ and therefore it stops.

From $t=\text{2}\text{ s}$ to $t=\text{4}\text{ s}$ the following happens:

The ball moves from a velocity of $\text{0}$ $\text{m·s $^{-1}$}$ to $\text{19.6}$ $\text{m·s $^{-1}$}$ in the opposite direction to the original motion.

If we assume that the ball is hit straight up in the air (and we take upwards as positive), it reaches its maximum height at $t= \text{2}\text{ s}$, stops, and falls back to the Earth to reach the ground at $t= \text{4}\text{ s}$.

Step 2: Draw the position-time graph

To draw this graph, we need to determine the displacements at $t = \text{2}\text{ s}$ and $t = \text{4}\text{ s}$.

At $t = \text{2}\text{ s}$:

The displacement is equal to the area under the graph:

Area under graph = Area of triangle

$\text{Area}=\frac{1}{2}bh$

$\text{Area}=\frac{1}{2}\times 2\times \text{19.6}$

Position = $\text{19.6}$ $\text{m}$

At $t = \text{4}\text{ s}$:

The displacement is equal to the area under the whole graph (top and bottom). Remember that an area under the time line must be subtracted:

Area under graph = Area of triangle 1 + Area of triangle 2

$\text{Area}=\frac{1}{2}bh+\frac{1}{2}bh$

$\text{Area}=\left(\frac{1}{2}\times 2\times \text{19.6}\right)+\left(\frac{1}{2}\times 2\times \left( -\text{19.6}\right)\right)$

$\text{Area}=19.6-19.6$

Displacement = $\text{0}$ $\text{m}$

The position vs time graph for motion at constant acceleration is a curve. The graph will look like this:

Step 3: Draw the acceleration versus time graph

To draw the acceleration vs time graph, we need to know what the acceleration is. The velocity versus time graph is a straight line which means that the acceleration is constant. The gradient of the line will give the acceleration.

The line has a negative slope (goes down towards the left) which means that the acceleration has a negative value.

Calculate the gradient of the line:

\begin{align*} \text{gradient}=\frac{\Delta \vec{v}}{\Delta t} \\ \text{gradient}=\frac{0 -\text{19,6}}{2-0} \\ \text{gradient}=\frac{ -\text{19,6}}{2} \\ \text{gradient}= -\text{9,8} \end{align*}

Therefore the acceleration = $\text{9.8}$ $\text{m·s $^{-2}$}$ downwards.