<p>The following example illustrates how to draw graphs of projectile motion.</p>
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<h2>Example: Drawing Graphs of Projectile Motion</h2>
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<h3>Question</h3>
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<p>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 $^{-1}$}$. The ball travels upwards to a maximum height, and then falls to the ground. Draw graphs of position vs time, velocity vs time and acceleration vs time. Choose upwards as the positive direction.</p>
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<h3>Step 1: Determine what is required</h3>
<p>We are required to draw graphs of:</p>
<ol>
<li>$\vec{x}$ vs $t$</li>
<li>$\vec{v}$ vs $t$</li>
<li>$\vec{a}$ vs $t$</li>
</ol>
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<h3>Step 2: Determine how to approach the problem</h3>
<p>There are two parts to the motion of the ball:</p>
<ol>
<li>ball travelling upwards from the building</li>
<li>ball falling to the ground</li>
</ol>
<p>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.</p>
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<h3>Step 3: Find the height and the time taken for the first motion</h3>
<p>For the first part of the motion we have:</p>
<ul>
<li>$\vec{v}_{i}= \text{+4.9}\text{ m·s $^{-1}$}$</li>
<li>$\vec{v}_{f}= \text{0}\text{ m·s $^{-1}$}$</li>
<li>$\vec{g}= -\text{9.8}\text{ m·s $^{-2}$}$</li>
</ul>
<img alt="65f988a6c607e8ca1f6046b2a9c5c1af.png" src="https://data.ulearngo.com/assets/a2cec3d0-5280-4b8f-afc0-e8c9e608da13"/>
<p>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.</p>

\begin{align*} \vec{v}_{f}^{2}&amp; = \vec{v}_{i}^{2}+2\vec{g}\Delta \vec{x} \\ {\left(0\right)}^{2}&amp; = {\left( \text{4,9}\right)}^{2}+2\times \left(-\text{9,8}\right)\times \Delta \vec{x} \\ \text{19,6}\Delta x&amp; = {\left(\text{4,9}\right)}^{2} \\ \Delta \vec{x}&amp; = + \text{1,225}\text{ m} \\ h &amp; = \Delta x = \text{1,225}\text{ m} \end{align*}\begin{align*} \vec{v}_{f}&amp; = \vec{v}_{i}+\vec{g}t \\ 0&amp; = \text{4,9}+\left( -\text{9,8}\right)\times t \\ \text{9,8}t&amp; = \text{4,9}\\ t&amp; = \text{0,5}\text{ s} \end{align*}</div>
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<h3>Step 4: Find the height and the time taken for the second motion</h3>
<img alt="251f56e4e973568b0a601c557e01c653.png" src="https://data.ulearngo.com/assets/ad9778eb-7531-4419-9bfe-b1eb1175b65d"/>
<p>For the second part of the motion we have:</p>
<ul>
<li>$\vec{v}_{i}=\text{0}\text{ m·s$^{-1}$}$</li>
<li>$\Delta \vec{x}=-\left(\text{20}+\text{1.225}\right) \text{m}$</li>
<li>$\vec{g}= -\text{9.8}\text{ m·s $^{-2}$}$</li>
</ul>
<p>Therefore we can use $\Delta \vec{x}=\vec{v}_{i}t+\frac{1}{2}\vec{g}{t}^{2}$ to solve for the time.</p>

\begin{align*} \Delta \vec{x}&amp; = \vec{v}_{i}t+\frac{1}{2}\vec{g}{t}^{2} \\ -\left(\text{20}+\text{1,225}\right)&amp; = \left(0\right)\times t+\frac{1}{2}\times \left( -\text{9,8}\right)\times {t}^{2} \\ -\text{21,225}&amp; = 0 -\text{4,9}{t}^{2} \\ {t}^{2}&amp; = \text{4,33163}... \\ t&amp; = \text{2,08125}\text{ s} \end{align*}<img alt="921f03aba0f036866d00b76d79a8ccef.png" src="https://data.ulearngo.com/assets/2d547566-6130-4a6a-8b8b-cd0101847c67"/></div>
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<h3>Step 5: Graph of position vs time</h3>
<p>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 $t=\text{0.5}+\text{2.08}=\text{2.58}~\text{s}$).</p>
<img alt="cd9db29a930a75061c0c6b40bf6f66da.png" src="https://data.ulearngo.com/assets/8b9c91df-9440-4ca7-bc90-ad7484021ea6"/></div>
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<h3>Step 6: Graph of velocity vs time</h3>
<p>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:</p>

\begin{align*} \vec{v}_{f}&amp; = \vec{v}_{i}+\vec{g}t \\ &amp; = 0+\left( -\text{9,8}\right)\left( \text{2,08}...\right) \\ &amp; = -\text{20,396}... \text{m}·{\text{s}}^{-1} \end{align*}<img alt="3d80e1000391e36aa4bf19ebae99f693.png" src="https://data.ulearngo.com/assets/60ba14d0-87c5-4b08-bc5f-0fd3b49501c3"/></div>
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<h3>Step 7: Graph of $a$ vs $t$</h3>
<p>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:</p>
<img alt="29155da30cd21c2f5c4f449bb8f5a8b2.png" src="https://data.ulearngo.com/assets/b208b589-cf5d-4e46-bef0-1a3db6a868c0"/></div>
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Drawing Graphs of Projectile Motion

Physics is a science discipline concerned with the nature and properties of matter and energy. Physics topics include mechanics, heat, light and radiation, sound, electricity, magnetism, the structure of atoms, and so on.

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Learn about the basic principles of motion including reference frames, position, displacement, distance, speed, velocity, acceleration, stationary objects, graphs of motion, projectile motion, and equations of motion with practical examples and applications.


Drawing Graphs of Projectile Motion

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