<p>The following example illustrates how to find the height of a projectile.</p>
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<h2>Example: Height of a Projectile</h2>
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<h3>Question</h3>
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<p>A cricketer hits a cricket ball so that it goes vertically upwards. If the ball takes $\text{10}$ $\text{s}$ to return to the initial height, determine its maximum height above the initial position.</p>
<img alt="" src="https://data.ulearngo.com/assets/adbb3fbd-2352-4c8a-92dc-65e6da97f832"/></div>
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<h3>Step 1: Identify what is required and what is given</h3>
<p>We need to find how high the ball goes. We know that it takes $\text{10}$ $\text{s}$ to go up and down. We do not know what the initial velocity ($\vec{v}_{i}$) of the ball is.</p>
<p>After the batsman strikes the ball it goes directly upwards and the only force that will be acting on the ball will be the gravitational force. This information is represented in the photograph below:</p>
<p><img alt="1cf1f27422ef46559be6eb2128fef1cd.png" src="https://data.ulearngo.com/assets/0f983e0c-abc8-416a-8879-c1a95b356c1c"/></p>
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<h3>Step 2: Determine how to approach the problem</h3>
<p>For a problem like this it is useful to divide the motion into two parts. The first part of the motion consists of the upward motion of the ball with an initial velocity ($\vec{v}_{i}$) and a final velocity ($\vec{v}_{f}=0$) at the maximum height. The second part of the motion consists of the downward motion of the ball with an initial velocity ($\vec{v}_{i}=0$) and final velocity ($\vec{v}_{f}$) which is not yet known.</p>
<img alt="67e872447f0471c4ce549b1573a76991.png" src="https://data.ulearngo.com/assets/a638ab55-d228-4ac8-945a-6def0ff450b4"/>
<p><strong>Choose down as positive</strong>. We know that at the maximum height, the velocity of the ball is $\text{0}$ $\text{m·s $^{-1}$}$. We also know that the ball takes the same time to reach its maximum height as it takes to travel from its maximum height to the initial position, due to time symmetry. The time taken is half the total time. Therefore, we have the following information for the second (downward) part of the motion of the ball:</p>
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<li>$t=\text{5}\text{ s}$ (half of the total time)</li>
<li>$\vec{v}_{\mathrm{top}}=\text{0}\text{ m·s$^{-1}$}$</li>
<li>$\vec{g}= \text{9.8}\text{ m·s $^{-2}$}$ downwards</li>
<li>$\Delta \vec{x}=~?$</li>
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<h3>Step 3: Find an appropriate equation to use</h3>
<p>We do not know the final velocity of the ball coming down. We need to choose an equation that does not have $\vec{v}_{f}$ in it. We can use the following equation to solve for $\Delta \vec{x}$:</p>

\[\Delta \vec{x}=\vec{v}_{i}t+\frac{1}{2}\vec{g}{t}^{2}\]</div>
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<h3>Step 4: Substitute values and find the height</h3>

\begin{align*} \Delta \vec{x}&amp; = \left( \text{0}\right)\left( \text{5}\right)+\frac{ \text{1}}{ \text{2}}\left( \text{9,8}\right){\left( \text{5}\right)}^{2} \\ \Delta \vec{x}&amp; = \text{122,5}\text{ m}~\text{downwards}\end{align*}

<p>In the second motion, the displacement of the ball is $\text{122.5}$ $\text{m}$ downwards. This means that the height was $h = \text{122.5}\text{ m}$.</p>
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<h3>Step 5: Write the final answer</h3>
<p>The ball reaches a maximum height of $\text{122.5}$ $\text{m}$.</p>
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Height of a Projectile Example

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.

Physics

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.


Height of a Projectile Example

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