Drawing Graphs of Projectile Motion

The following example illustrates how to draw graphs of projectile motion.

Example: Drawing Graphs of Projectile Motion

Question

Stanley is standing on a balcony $\text{20}$ $\text{m}$ above the ground. Stanley tosses a rubber ball upwards with an initial velocity of $\text{4.9}$ $\text{m·s $^{-1}$}$. The ball travels upwards to a maximum height, and then falls to the ground. Draw graphs of position vs time, velocity vs time and acceleration vs time. Choose upwards as the positive direction.

Step 1: Determine what is required

We are required to draw graphs of:

$\vec{x}$ vs $t$
$\vec{v}$ vs $t$
$\vec{a}$ vs $t$

Step 2: Determine how to approach the problem

There are two parts to the motion of the ball:

ball travelling upwards from the building
ball falling to the ground

We examine each of these parts separately. To be able to draw the graphs, we need to determine the time taken and displacement for each of the motions.

Step 3: Find the height and the time taken for the first motion

For the first part of the motion we have:

$\vec{v}_{i}= \text{+4.9}\text{ m·s $^{-1}$}$
$\vec{v}_{f}= \text{0}\text{ m·s $^{-1}$}$
$\vec{g}= -\text{9.8}\text{ m·s $^{-2}$}$

Therefore we can use $\vec{v}_{f}^{2}=\vec{v}_{i}^{2}+2\vec{g}\Delta \vec{x}$ to solve for the height and $\vec{v}_{f}=\vec{v}_{i}+\vec{g}t$ to solve for the time.

\begin{align*} \vec{v}_{f}^{2}& = \vec{v}_{i}^{2}+2\vec{g}\Delta \vec{x} \\ {\left(0\right)}^{2}& = {\left( \text{4,9}\right)}^{2}+2\times \left(-\text{9,8}\right)\times \Delta \vec{x} \\ \text{19,6}\Delta x& = {\left(\text{4,9}\right)}^{2} \\ \Delta \vec{x}& = + \text{1,225}\text{ m} \\ h & = \Delta x = \text{1,225}\text{ m} \end{align*}\begin{align*} \vec{v}_{f}& = \vec{v}_{i}+\vec{g}t \\ 0& = \text{4,9}+\left( -\text{9,8}\right)\times t \\ \text{9,8}t& = \text{4,9}\\ t& = \text{0,5}\text{ s} \end{align*}

Step 4: Find the height and the time taken for the second motion

For the second part of the motion we have:

$\vec{v}_{i}=\text{0}\text{ m·s$^{-1}$}$
$\Delta \vec{x}=-\left(\text{20}+\text{1.225}\right) \text{m}$
$\vec{g}= -\text{9.8}\text{ m·s $^{-2}$}$

Therefore we can use $\Delta \vec{x}=\vec{v}_{i}t+\frac{1}{2}\vec{g}{t}^{2}$ to solve for the time.

\begin{align*} \Delta \vec{x}& = \vec{v}_{i}t+\frac{1}{2}\vec{g}{t}^{2} \\ -\left(\text{20}+\text{1,225}\right)& = \left(0\right)\times t+\frac{1}{2}\times \left( -\text{9,8}\right)\times {t}^{2} \\ -\text{21,225}& = 0 -\text{4,9}{t}^{2} \\ {t}^{2}& = \text{4,33163}... \\ t& = \text{2,08125}\text{ s} \end{align*}

Step 5: Graph of position vs time

The ball starts from a position of $\text{20}$ $\text{m}$ (at t = $\text{0}$ $\text{s}$) from the ground and moves upwards until it reaches $\text{20}$+$\text{1.225}$ $\text{m}$ (at $t=\text{0.5}\text{ s}$). It then falls back to $\text{20}$ $\text{m}$ (at $t=\text{0.5}+\text{0.5}= \text{1.0}\text{s}$) and then falls to the ground, $\Delta \vec{x}=\text{0}\text{ m}$ (at $t=\text{0.5}+\text{2.08}=\text{2.58}~\text{s}$).

Step 6: Graph of velocity vs time

The ball starts off with a velocity of $\text{+4.9}$ $\text{m·s $^{-1}$}$ at t = $\text{0}$ $\text{s}$, it then reaches a velocity of $\text{0}$ $\text{m·s $^{-1}$}$ at t = $\text{0.5}$ $\text{s}$. It stops and falls back to the Earth. At t = $\text{1.0}$ $\text{s}$ (i.e. after a further $\text{0.5}$ $\text{s}$) it has a velocity of $-\text{4.9}$ $\text{m·s $^{-1}$}$. This is the same as the initial upwards velocity but it is downwards. It carries on at constant acceleration until $t=\text{2.58}\text{ s}$. In other words, the velocity graph will be a straight line. The final velocity of the ball can be calculated as follows:

\begin{align*} \vec{v}_{f}& = \vec{v}_{i}+\vec{g}t \\ & = 0+\left( -\text{9,8}\right)\left( \text{2,08}...\right) \\ & = -\text{20,396}... \text{m}·{\text{s}}^{-1} \end{align*}

Step 7: Graph of $a$ vs $t$

We chose upwards to be positive. The acceleration of the ball is downward, $\vec{g}= \text{9.8}\text{ m·s $^{-2}$}$ downwards. Because the acceleration is constant throughout the motion, the graph looks like this: