A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.
Solution:
Part A
Consider Figure A.
We are interested in two positions. Position 1 is where the coin is dropped. At this position, the coin is 300 m above the ground, the time is 0 s, and the velocity is 10.0 m/s upward.
Position 2 is the highest point of the coin reaches. At this position, the velocity is equal to 0 m/s.
Position 1 is the initial position and position 2 is the final position. Solve for the value of y2.
\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 +2a \Delta y \\
\Delta y & = \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} \\
y_2 - y_1 & = \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} \\
y_2& = \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} +y_1 \\
y_2 & = \frac{\left( 0 \ \text{m/s} \right)^2-\left( 10.0\ \text{m/s} \right)^2}{2\left( -9.81 \ \text{m/s}^2 \right)}+300\ \text{m}
\\
y_2 & =305 \ \text{m} \ \qquad {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}\therefore The maximum height reached by the coin is about 305 meters from the ground.
Part B
We do not know the position 4 seconds after the coin has been released, the answer can be above or below the initial point. We can actually use one of the kinematical equations to solve for the final position given the time. Here, the initial position is the point of release and the final position is the point of interest at 4.00 seconds after release.
\begin{align*}
\Delta y & = v_{y_1}t+\frac{1}{2}at^2 \\
y_2 - y_1 & = v_{y_1}t+\frac{1}{2}at^2 \\
y_2 & = y_1 + v_{y_1}t+\frac{1}{2}at^2 \\
y_2 & = 300\ \text{m} +\left( 10\ \text{m/s} \right)\left( 4.00\ \text{s} \right)+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 4.00\ \text{s} \right)^2\\
y_2 & = 261.52\ \text{m} \\
y_2 & = 262\ \text{m}\ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}\therefore The coin is at a height of 262 meters above the ground 4.00 seconds after release. That is, the coin is already dropping and it is already below the release point.
Solving for the velocity 4.00 seconds after release considering the same initial and final position.
\begin{align*}
v_{y_2} & = v_{y_1}+at \\
v_{y_2} & = 10\ \text{m/s} + \left( -9.81\ \text{m/s}^2 \right)\left( 4.00 \ \text{s} \right) \\
v_{y_2} & = -29.2\ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}\therefore The coin has a velocity of 29.2 m/s directed downward 4.00 seconds after it is released. This confirms that the coin is indeed moving downwards at this point.
Part C
Considering figure C, we have two positions. Position 1 is the point of release 300 m above the ground with a velocity of 10 m/s upward. This is time 0 s.
The second position is at the ground where y=0 m. We are interested at the time in this position.
Considering position 1 as the initial position and position 2 as the final position.
\begin{align*}
\Delta y & = v_{y_1} \Delta t+\frac{1}{2}a\left( \Delta t \right)^2 \\
y_2-y_1 & = v_{y_1}t+\frac{1}{2}at^2 \\
0\ \text{m}-300\ \text{m} & = \left( 10 \ \text{m/s} \right)t+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)t^2 \\
-300 & = 10t-4.905t^2 \\
4.905t^2-10t-300 & = 0 \\
\end{align*}Solve for the value of t using the quadratic formula with a=4.905, b=-10, and c=-300.
\begin{align*}
t & = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\\
t & = \frac{-\left( -10 \right) \pm \sqrt{\left( 10 \right)^2-4\left( 4.905 \right)\left( -300 \right)}}{2\left( 4.905 \right)}\\
t & = 8.91 \ \text{s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}\therefore The time is about 8.91 seconds before the coin hits the ground.
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