<p>The following example further illustrates projectile motion using a hot-air balloon.</p>
<div class="ns-school-emp">
<h2>Example: Hot Air Balloon</h2>
<div class="ns-school-defn">
<h3>Question</h3>
<div>
<p>A hot-air balloon is moving vertically upwards at a constant speed. A camera is accidentally dropped from the balloon at a height of $\text{92.4}$ $\text{m}$ as shown in the diagram below. The camera strikes the ground after $\text{6}$ $\text{s}$. Ignore the effects of friction.</p>
<img alt="0ae046ec701be47ab2d8c15ac91b9548.png" src="https://data.ulearngo.com/assets/374682cf-2419-4d3d-bfb3-9344c022d6e6"/>
<ol>
<li>
<p>At the instant the camera is dropped, it moves upwards. Give a reason for this observation.</p>
</li>
<li>
<p>Calculate the speed $v_i$ at which the balloon is rising when the camera is dropped.</p>
</li>
<li>
<p>Draw a sketch graph of velocity versus time for the entire motion of the camera.</p>
<p>Indicate the following on the graph:</p>
<ul>
<li>Initial velocity</li>
<li>Time at which it reaches the ground</li>
</ul>
</li>
<li>
<p>If a jogger, $\text{10}$ $\text{m}$ away from point <strong>P</strong> as shown in the above diagram and running at a constant speed of $\text{2}$ $\text{m·s$^{-1}$}$, sees the camera at the same instant it starts falling from the balloon, will he be able to catch the camera before it strikes the ground?</p>
<p>Use a calculation to show how you arrived at the answer.</p>
</li>
</ol>
</div>
<h3><strong>Question 1 Solution</strong></h3>
<p>The initial velocity / speed of the camera is the same (as that of the balloon).</p>
<h3><strong>Question 2 Solution</strong></h3>
<p><strong>Taking downwards as the positive direction:</strong></p>

\begin{align*} \Delta y &amp;= v_i \Delta t + \frac{1}{2} a \Delta t^2 \\ \therefore \text{92,4} &amp;= v_i (\text{6}) + \frac{1}{2} (\text{9,8}) (\text{6})^2 \\ \therefore v_i &amp;= -\text{14 m}\cdot\text{s}^{-1} \\ \therefore v_i &amp;= \text{14 m}\cdot\text{s}^{-1} \end{align*}

<p><strong>Taking downwards as the negative direction:</strong></p>

\begin{align*} \Delta y &amp;= v_i \Delta t + \frac{1}{2} a \Delta t^2 \\ \therefore -\text{92,4} &amp;= v_i (\text{6}) + \frac{1}{2} (-\text{9,8}) (\text{6})^2 \\ \therefore v_i &amp;= \text{14 m}\cdot\text{s}^{-1} \end{align*}

<h3><strong>Question 3 Solution</strong></h3>
<p><strong>Taking downwards as the positive direction:</strong></p>
<img alt="02a15ba4712290dfce203490528ba9e0.png" src="https://data.ulearngo.com/assets/02159b19-fd5b-4f39-80a7-1f995c1c912f"/>
<p><strong>Criteria for graph:</strong></p>
<ul>
<li>Correct shape as shown. (Straight line with gradient.)</li>
<li>Graph starts at $v = \text{14 m}\cdot\text{s}^{-1}\ /\ v_i$ at $t = \text{0 s}.$</li>
<li>Graph extends below $t$ axis until $t = \text{6 s}.$</li>
<li>Section of graph below $t$ axis is longer than section above $t$ axis.</li>
</ul>
<p><strong>Taking downwards as the negative direction:</strong></p>
<img alt="ad3283d5be696c8dd87d8fd7cf7d043f.png" src="https://data.ulearngo.com/assets/1d1f8faf-55ef-48dd-a423-a5d0aaaad4cd"/>
<p><strong>Criteria for graph:</strong></p>
<ul>
<li>Correct shape as shown. (Straight line with gradient.)</li>
<li>Graph starts at $v = -\text{14 m}\cdot\text{s}^{-1}\ /\ v_i$ at $t = \text{0 s}.$</li>
<li>Graph extends above $t$ axis until $t = \text{6 s}.$</li>
<li>Section of graph above $t$ axis is longer than section below $t$ axis.</li>
</ul>
<h3><strong>Question 4 Solution</strong></h3>
<p><strong>Option 1:</strong></p>

\begin{align*} \Delta x &amp;= v \Delta t \\ \therefore \text{10} &amp;= (\text{2}) \Delta t \\ \therefore \Delta t &amp;= \text{5 s} \end{align*}

<p>Yes, he will catch the camera since the time is less than $\text{6}$ $\text{s}$.</p>
<p><strong>Option 2:</strong></p>

\begin{align*} \Delta x &amp; = v \Delta t \\ \therefore &amp; = (\text{2})(\text{6}) \\ \therefore &amp; = \text{12 m} \end{align*}

<p>Yes, he will catch the camera since the distance covered is greater than $\text{10}$ $\text{m}$.</p>
<p><strong>Option 3:</strong></p>

\begin{align*} \Delta x &amp; = v_i \Delta t + \frac{1}{2} a \Delta t^2 \\ \therefore \text{10} &amp; = (\text{2})\Delta t + \frac{1}{2}(\text{0})\Delta t^2 \\ \therefore \Delta t &amp; = \text{5 t} \end{align*}

<p>Yes, he will catch the camera since the time is less than $\text{6}$ $\text{s}$.</p>
<p><strong>Option 4:</strong></p>

\begin{align*} \Delta x &amp; = \left( \frac{v_i+v_f}{2} \right ) \Delta t \\ \therefore \text{10} &amp; = \left( \frac{\text{2}+\text{2}}{2} \right ) \Delta t \\ \therefore \Delta t &amp; = \text{5 t} \end{align*}

<p>Yes, he will catch the camera since the time is less than $\text{6}$ $\text{s}$.</p>
<p><strong>Option 5:</strong></p>

\begin{align*} \Delta x &amp; = \left( \frac{v_i+v_f}{2} \right ) \Delta t \\ &amp; = \left( \frac{\text{2}+\text{2}}{2} \right )(\text{6}) \\ \therefore \Delta x &amp; = \text{12 m} \end{align*}

<p>Yes, he will catch the camera since the distance covered is greater than $\text{10}$ $\text{m}$.</p>
</div>
</div>

1c4ddb66-ad36-4f81-90e7-4629f497d0b4

Hot Air Balloon

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.


Hot Air Balloon

Track Your Learning Progress