A hot air balloon ascends at \( 4 \, \text{m/s} \) for \( 6 \, \text{s} \), then accelerates at \( 1.5
\,
\text{m/s}^2 \) for \( 4 \, \text{s} \). A stone is dropped at this instant. How long does it take to
reach the ground if the initial height was \( 30 \, \text{m} \)? (Take \( g = 10 \, \text{m/s}^2 \))
Phase 1: \( h_1 = 4 \cdot 6 = 24 \, \text{m} \).
Phase 2: \( v = 4 + 1.5 \cdot 4 = 10 \, \text{m/s} \), \( h_2 = 4 \cdot 4 + \frac{1}{2} \cdot 1.5 \cdot
(4)^2 = 16 + 12 = 28 \, \text{m} \).
Total height = \( 30 + 24 + 28 = 82 \, \text{m} \), stone’s initial velocity = \( 10 \, \text{m/s} \)
upward.
\( -82 = 10 t - 5 t^2 \Rightarrow 5 t^2 - 10 t - 82 = 0 \).
Solve: \( t = \frac{10 \pm \sqrt{100 + 1640}}{10} = \frac{10 \pm 41.23}{10} \), \( t = 5.12 \, \text{s}
\)
(positive root).