# College Physics 2.57 – A coin dropped from a hot-air balloon

## Solution:

### Part A

For the coin, we are given the following:  $v_o=+10\:m/s;\:a=-9.80\:m/s^2;\:v=0\:m/s\:\left(at\:top\:of\:ascent\right)$

The maximum height reached can be solved using the formula

$v^2=\left(v_o\right)^2+2a\Delta y$

Solve for Δy in terms of the other variables

$\Delta y=\frac{v^2-\left(v_o\right)^2}{2a}$

Now, substitute the given values

$\Delta y=\frac{\left(0\:m/s\right)^2-\left(10\:m/s\right)^2}{2\left(-9.80\:m/s\right)^2}$

$\Delta y=5.10\:m$

From the moment the coin is dropped to the top of the ascent, the change in position of the coin is +5.10 m. Therefore, the maximum height is

$y_{max}=300\:m+5.10\:m$

$y_{max}=305.10\:m$

### Part B

The position of the coin after 4 seconds can be solved using the formula

$\Delta y=v_ot+\frac{1}{2}at^2$

Substitute the given values

$y=\left(10\:m/s\right)\left(4\:s\right)+\frac{1}{2}\left(-9.80\:m/s^2\right)\left(4\:s\right)^2$

$y=-38.4\:m$

The change in position of the coin is -38.4 m. Therefore, the poisiton of the coin with respect to the ground is

$y_{_{\left(4\:s\right)}}=300\:m-38.4\:m$

$y_{_{\left(4\:s\right)}}=261.6\:m$

The velocity after 4 seconds is given by the formula

$v=v_o+at$

Substitute the given values

$v=10\:m/s+\left(-9.8\:m/s^2\right)\left(4.00\:s\right)$

$v=-29.2\:m/s$

The velocity is 29.2 m/s, downward.

### Part C

The time it takes for the coin to hit the ground can be solved using the formula

$\Delta y=v_ot+\frac{1}{2}at^2$

Substitute the given values

$300\:m=\left(10.0\:m/s\right)t+\frac{1}{2}\left(-9.80\:m/s^2\right)t^2$

Simplify and rearrange the equation to form a quadratic equation

$-4.90t^2+10t-300=0$

Now, we can solve for time, t, using the quadratic formula

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$t=\frac{-\left(10\right)\pm \sqrt{\left(10\right)^2-4\left(-4.9\right)\left(300\right)}}{2\left(-4.9\right)}$

$t=8.91\:s$

The time of flight is 8.91 seconds.