<p>The following example shows calculations involving projectile motion.</p>
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<h2>Example: Projectile Motion</h2>
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<h3>Question</h3>
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<p>A ball is thrown vertically upwards with an initial velocity of $\text{10}$ $\text{m·s $^{-1}$}$. A photograph (shown) is taken when the ball is $\text{1.5}$ $\text{m}$ above the point of release.</p>
<img alt="" src="https://data.ulearngo.com/assets/9b81841d-422c-44e4-8e15-69fc9987877d"/>
<ol>
<li>
<p>Determine the maximum height above the thrower's hand reached by the ball.</p>
</li>
<li>
<p>Determine the time it takes the ball to reach its maximum height.</p>
</li>
<li>
<p>Determine the time(s) when the ball will be at the position shown in the photograph.</p>
</li>
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<h3>Step 1: Identify what is required and what is given</h3>
<p>We are required to determine the maximum height reached by the ball and how long it takes to reach this height. We are given the initial velocity $\vec{v}_{i}=\text{10}\text{ m·s$^{-1}$}$ upwards and the acceleration due to gravity $\vec{g}= \text{9.8}\text{ m·s $^{-2}$}$ downwards. We also need to determine when the ball is at the position shown in the photograph.</p>
<p>The ball is thrown straight up and will reach a maximum height, at which point its velocity will be zero, before beginning to fall downwards. The height of the ball in the photograph is given as $\text{1.5}$ $\text{m}$ above the initial point. Adding the information to the picture helps us understand what is going on more effectively:</p>
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<h3>Step 2: Determine how to approach the problem</h3>
<p>Even though there are a number of questions to answer, the best approach is to work through them one at a time.</p>
<p>First, as with rectilinear motion in Grade 10, select a positive direction. We choose <strong>down as positive</strong> and remember to <strong>keep this sign convention</strong> for the whole problem. This will allow us to analyse the information given and determine directions and signs. We know that at the maximum height the velocity of the ball is $\text{0}$ $\text{m·s $^{-1}$}$. We therefore have the following:</p>
<ul>
<li>$\vec{v}_{i}= -\text{10}\text{ m·s $^{-1}$}$(it is negative because we chose downwards as positive)</li>
<li>$\vec{v}_{f}= \text{0}\text{ m·s $^{-1}$}$</li>
<li>$\vec{g}= \text{9.8}\text{ m·s $^{-2}$}$</li>
</ul>
<p>We know that the position shown ($\text{1.5}$ $\text{m}$ above the point of release) will be passed as the ball rises to its maximum height and then again as the ball falls downwards.</p>
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<h3>Step 3: Identify the appropriate equation to determine the maximum height</h3>
<p>We can use $\vec{v}_{f}^{2}=\vec{v}_{i}^{2}+2\vec{g}\Delta \vec{x}$ to solve for the height.</p>
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<h3>Step 4: Substitute the values into the equation and find the maximum height</h3>

\begin{align*} \vec{v}_{f}^{2}&amp; = \vec{v}_{i}^{2}+2\vec{g}\Delta \vec{x} \\ {\left(0\right)}^{2}&amp; = {\left( -\text{10}\right)}^{2}+\left(2\right)\left( \text{9,8}\right)\left(\Delta x\right) \\ -\text{100}&amp; = \text{19,6}\Delta x \\ \Delta x&amp; = -\text{5,102}\text{ m}\end{align*}

<p>The value for the displacement will be negative because the displacement is upwards and we have chosen downward as positive (and upward as negative). The maximum height will be a positive number, $h_m= \text{5.10}\text{ m}$.</p>
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<h3>Step 5: Identify the appropriate equation to determine the time to reach maximum height</h3>
<p>We know the values of $\vec{v}_i$, $\vec{v}_f$ and $\vec{g}$ so we can use: $\vec{v}_{f}=\vec{v}_{i}+\vec{g}t$ to solve for the time.</p>
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<h3>Step 6: Substitute the values in and find the time to reach maximum height</h3>

\begin{align*} \vec{v}_{f}&amp; = \vec{v}_{i}+\vec{g}t \\ \text{0}&amp; = -\text{10}+ \text{9,8}t \\ \text{10}&amp; = \text{9,8}t \\ t&amp; = \text{1,02}\text{ s}\end{align*}</div>
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<h3>Step 7: Determine the times at which the ball is in the position in the photograph</h3>
<p>We know the displacement, initial velocity and the acceleration of the ball. We also expect to get two values as we know the ball will pass through the point twice.</p>
<p>We substitute our values into the equation $\Delta \vec{x} = \vec{v}_{i}t+\frac{1}{2}\vec{g}{t}^{2}$, which is a quadratic equation in time, so we expect to find two different solutions if we solve for time.</p>
<p>It is important to notice that we can use the value $-\text{1.5}$ $\text{m}$ as the displacement above the initial position.</p>

$\begin{align*} <br/>
 \Delta \vec{x} &amp; = {v}_{i}t+\frac{1}{2}\vec{g}{t}^{2} \\ <br/>
 \frac{1}{2}\vec{g}{t}^{2} + \vec{v}_{i}t - \Delta \vec{x} &amp; = \text{0}\\ <br/>
 \frac{1}{2}( \text{9.8})t^2 + ( -\text{10})t - ( -\text{1.5}) &amp; = \text{0}\\ <br/>
 \underbrace{( \text{4,9})}_{a}\underbrace{t^2}_{x^2} + \underbrace{( -\text{10})}_{b}\underbrace{t}_{x}+\underbrace{(\text{1.5})}_{c} &amp; = 0 \\ <br/>
 t &amp;= \frac{-b\pm\sqrt{b^2- \text{4}ac}}{ \text{2}a} \\ <br/>
 t &amp;= \frac{-( -\text{10})\pm\sqrt{( -\text{10})^2- \text{4}( \text{4.9})( \text{1.5})}}{ \text{2}( \text{4.9})} \\ <br/>
 t &amp;= \frac{ \text{10}\pm\sqrt{ \text{70.6}}}{ \text{9.8}} \\ t &amp;= \frac{ \text{10}\pm \text{8.40238}}{ \text{9.8}} \\ <br/>
 t &amp;= \text{1.02041}\pm \text{0.857386}\\ <br/>
 t = \text{0.16}\text{ s}\ &amp; \text{or}\ t = \text{1.88}\text{ s}<br/>
 \end{align*}$</div>
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<h3>Step 8: Write the final answer</h3>
<p>The ball:</p>
<ol>
<li>reaches a maximum height of $\text{5.10}$ $\text{m}$;</li>
<li>takes $\text{1.02}$ $\text{s}$ to reach the top; and</li>
<li>passes through the point in the photograph at $t$= $\text{0.16}$ $\text{s}$ on the way up and t = $\text{1.88}$ $\text{s}$ on the way down.</li>
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<p>The use of a sign convention should not affect your result. As an exercise, repeat the worked example using the opposite sign convention to verify that you do get the same result.</p>

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Projectile Motion Example

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Projectile Motion Example

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