A ballast bag is dropped from a balloon that is $300 \mathrm{~m}$ above the ground and rising at $13 \mathrm{~m} / \mathrm{s}$. For the bag, find $(a)$ the maximum height reached, $(b)$ its position and velocity $5.0 \mathrm{~s}$ after it is released, and $(c)$ the time at which it hits the ground.
The initial velocity of the bag when released is the same as that of the balloon, $13 \mathrm{~m} / \mathrm{s}$ upward. Choose $u p$ as positive and take $y=0$ at the point of release.
(a) At the highest point, $v_{f}=0$. From $v_{f y}^{2}=v_{i y}^{2}+2 a y$,
$$0=(13 \mathrm{~m} / \mathrm{s})^{2}+2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right) y \quad \text { or } \quad y=8.6 \mathrm{~m}$$
The maximum height is $300+8.6=308.6 \mathrm{~m}$ or $0.31 \mathrm{~km}$.
(b) Take the end point to be its position at $t=5.0 \mathrm{~s}$. Then, from $y=v_{i y} t+\frac{1}{2} a t^{2}$,
$$y=(13 \mathrm{~m} / \mathrm{s})(5.0 \mathrm{~s})+\frac{1}{2}\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(5.0 \mathrm{~s})^{2}=-57.6 \mathrm{~m} \text { or }-58 \mathrm{~m}$$
So its height is $300-58=242 \mathrm{~m}$. Also, from $v_{f y}=v_{i y}+a t$,
$$v_{f y}=13 \mathrm{~m} / \mathrm{s}+\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(5.0 \mathrm{~s})=-36 \mathrm{~m} / \mathrm{s}$$
It is on its way down with a velocity of $36 \mathrm{~m} / \mathrm{s}-$ DOWNWARD.
(c) Just as it hits the ground, the bag's displacement is $-300 \mathrm{~m}$. Then
$y=v_{i y} t+\frac{1}{2} a t^{2} \quad$ becomes $\quad-300 \mathrm{~m}=(13 \mathrm{~m} / \mathrm{s}) t+\frac{1}{2}\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}$
or $4.905 t^{2}-13 t-300=0$. The quadratic formula gives $t=9.3 \mathrm{~s}$ and $-6.6 \mathrm{~s}$. Only the positive time has physical meaning, so the required answer is $9.3 \mathrm{~s}$. as before.