Coin Dropped From a Hot-Air Balloon| Falling Objects| Kinematics| College Physics| Problem 2.57

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

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. 

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