College Physics by Openstax Chapter 2 Problem 57: Vertical motion analysis of a coin dropped from a rising hot-air balloon

Problem:

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.

College Physics Chapter 2 Problems

Problem 1	Problem 2	Problem 3	Problem 4	Problem 5
Problem 6	Problem 7	Problem 8	Problem 9	Problem 10
Problem 11	Problem 12	Problem 13	Problem 14	Problem 15
Problem 16	Problem 17	Problem 18	Problem 19	Problem 20
Problem 21	Problem 22	Problem 23	Problem 24	Problem 25
Problem 26	Problem 27	Problem 28	Problem 29	Problem 30
Problem 31	Problem 32	Problem 33	Problem 34	Problem 35
Problem 36	Problem 37	Problem 38	Problem 39	Problem 40
Problem 41	Problem 42	Problem 43	Problem 44	Problem 45
Problem 46	Problem 47	Problem 48	Problem 49	Problem 50
Problem 51	Problem 52	Problem 53	Problem 54	Problem 55
Problem 56	Problem 57	Problem 58	Problem 59	Problem 60
Problem 61	Problem 62	Problem 63	Problem 64	Problem 65
Problem 66