Hot Air Balloon
The following example further illustrates projectile motion using a hot-air balloon. A hot-air balloon is moving vertically upwards at a constant speed. A camera is accidentally dropped from the balloon at a height of 92.4 m as shown in the diagram in the lesson. The ...
The following example further illustrates projectile motion using a hot-air balloon.
Example: Hot Air Balloon
Question
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.
-
At the instant the camera is dropped, it moves upwards. Give a reason for this observation.
-
Calculate the speed \(v_i\) at which the balloon is rising when the camera is dropped.
-
Draw a sketch graph of velocity versus time for the entire motion of the camera.
Indicate the following on the graph:
- Initial velocity
- Time at which it reaches the ground
-
If a jogger, \(\text{10}\) \(\text{m}\) away from point P 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?
Use a calculation to show how you arrived at the answer.
Question 1 Solution
The initial velocity / speed of the camera is the same (as that of the balloon).
Question 2 Solution
Taking downwards as the positive direction:
\begin{align*} \Delta y &= v_i \Delta t + \frac{1}{2} a \Delta t^2 \\ \therefore \text{92,4} &= v_i (\text{6}) + \frac{1}{2} (\text{9,8}) (\text{6})^2 \\ \therefore v_i &= -\text{14 m}\cdot\text{s}^{-1} \\ \therefore v_i &= \text{14 m}\cdot\text{s}^{-1} \end{align*}Taking downwards as the negative direction:
\begin{align*} \Delta y &= v_i \Delta t + \frac{1}{2} a \Delta t^2 \\ \therefore -\text{92,4} &= v_i (\text{6}) + \frac{1}{2} (-\text{9,8}) (\text{6})^2 \\ \therefore v_i &= \text{14 m}\cdot\text{s}^{-1} \end{align*}Question 3 Solution
Taking downwards as the positive direction:
Criteria for graph:
- Correct shape as shown. (Straight line with gradient.)
- Graph starts at \(v = \text{14 m}\cdot\text{s}^{-1}\ /\ v_i\) at \(t = \text{0 s}.\)
- Graph extends below \(t\) axis until \(t = \text{6 s}.\)
- Section of graph below \(t\) axis is longer than section above \(t\) axis.
Taking downwards as the negative direction:
Criteria for graph:
- Correct shape as shown. (Straight line with gradient.)
- Graph starts at \(v = -\text{14 m}\cdot\text{s}^{-1}\ /\ v_i\) at \(t = \text{0 s}.\)
- Graph extends above \(t\) axis until \(t = \text{6 s}.\)
- Section of graph above \(t\) axis is longer than section below \(t\) axis.
Question 4 Solution
Option 1:
\begin{align*} \Delta x &= v \Delta t \\ \therefore \text{10} &= (\text{2}) \Delta t \\ \therefore \Delta t &= \text{5 s} \end{align*}Yes, he will catch the camera since the time is less than \(\text{6}\) \(\text{s}\).
Option 2:
\begin{align*} \Delta x & = v \Delta t \\ \therefore & = (\text{2})(\text{6}) \\ \therefore & = \text{12 m} \end{align*}Yes, he will catch the camera since the distance covered is greater than \(\text{10}\) \(\text{m}\).
Option 3:
\begin{align*} \Delta x & = v_i \Delta t + \frac{1}{2} a \Delta t^2 \\ \therefore \text{10} & = (\text{2})\Delta t + \frac{1}{2}(\text{0})\Delta t^2 \\ \therefore \Delta t & = \text{5 t} \end{align*}Yes, he will catch the camera since the time is less than \(\text{6}\) \(\text{s}\).
Option 4:
\begin{align*} \Delta x & = \left( \frac{v_i+v_f}{2} \right ) \Delta t \\ \therefore \text{10} & = \left( \frac{\text{2}+\text{2}}{2} \right ) \Delta t \\ \therefore \Delta t & = \text{5 t} \end{align*}Yes, he will catch the camera since the time is less than \(\text{6}\) \(\text{s}\).
Option 5:
\begin{align*} \Delta x & = \left( \frac{v_i+v_f}{2} \right ) \Delta t \\ & = \left( \frac{\text{2}+\text{2}}{2} \right )(\text{6}) \\ \therefore \Delta x & = \text{12 m} \end{align*}Yes, he will catch the camera since the distance covered is greater than \(\text{10}\) \(\text{m}\).
This lesson is part of:
One-Dimensional Motion