A balloon rises vertically at v_0 = 5 m/s . At a height of H…

A balloon rises vertically at $v_0 = 5\ \mathrm{m/s}$. At a height of $H = 30\ \mathrm{m}$ a load is released from it with zero velocity relative to the balloon. (a) At what height is the load $t_1 = 2\ \mathrm{s}$ after release? (b) How long does it take to reach the ground?

Explanation of Method

A load released from a rising balloon.

Step 1: Initial velocity of the load

"Zero velocity relative to the balloon" means that relative to the ground the load starts with the balloon's velocity: 5 m/s upward. From then on only gravity acts:

\[ y(t) = H + v_0 t - \tfrac12 g t^2 = 30 + 5t - 4.905\,t^2. \]

Step 2: (a) Height after 2 s

\[ y(2) = 30 + 10 - 19.62. \]

\[y(2\ \mathrm{s}) \approx 20.4\ \mathrm{m}\]

Step 3: (b) Time to the ground

\[ 4.905\,t^2 - 5t - 30 = 0 \;\Longrightarrow\; t = \frac{5 + \sqrt{25 + 588.6}}{9.81} = \frac{5 + 24.77}{9.81}. \]

\[t \approx 3.04\ \mathrm{s}\]

$h\ (\mathrm{m})$ $t\ (\mathrm{s})$ $0$ $0.51$ $2$ $3.04$ $20.4$ $30$ $40$ $\text{top: } 31.3\ \mathrm{m}$ $20.4\ \mathrm{m}\ \text{at}\ 2\ \mathrm{s}$ $\text{balloon}$ $\text{lands: } 3.04\ \mathrm{s}$

load
balloon

Height against time. The released load keeps the balloon's upward 5 m/s at first, so it rises another 1.3 m before falling. The balloon (dashed) keeps rising steadily.

Result

\[\boxed{\,y(2\ \mathrm{s}) \approx 20.4\ \mathrm{m}\,}\]

\[\boxed{\,t \approx 3.04\ \mathrm{s}\,}\]

Remark

Note. The load first rises for $5/9.81 \approx 0.51\ \mathrm{s}$ to 31.3 m before it starts falling. After 2 s the balloon is at 40 m, so the load is already 19.6 m below it; relative to the balloon the load simply "falls from rest", $\tfrac12 g t^2$.