Question

A hot air balloon is rising straight up at a constant speed of 7.0 m/s. When the balloon is 12.0m above the ground, a gun fires a pellet straight up from ground level with an initial speed of 30.0m/s. Along the paths of the balloon and the pellet, there are 2 places where each of them has the same altitude at the same time. How far above the ground are these 2 places?

Answer 1

Answer on Question #74006, Physics / Mechanics | Relativity

Question. A hot air balloon is rising straight up at a constant speed of $7.0 \, m/s$. When the balloon is $12.0 \, m$ above the ground, a gun fires a pellet straight up from ground level with an initial speed of $30.0 \, m/s$. Along the paths of the balloon and the pellet, there are 2 places where each of them has the same altitude at the same time. How far above the ground are these 2 places?

Given. $v = 7.0 \, m/s; h_0 = 12.0 \, m; u = 30.0 \, m/s$.

Find. $h_1, h_2 - ?$

Solution.

For a hot air balloon

$$h = h_0 + vt.$$

For a pellet

$$s = ut - \frac{gt^2}{2}.$$ $$h = s \rightarrow h_0 + vt = ut - \frac{gt^2}{2}$$ $$12 + 7t = 30t - \frac{9.8t^2}{2} \rightarrow 4.9t^2 - 23t + 12 = 0 \Rightarrow t_1 = 0.6 \, s; \, t_2 = 4.1 \, s.$$

Hence

$$h_1 = 12 + 7 \cdot 0.6 = 16.2 \, m;$$ $$h_2 = 12 + 7 \cdot 4.1 = 40.7 \, m;$$

Answer. $h_1 = 16.2 \, m; h_2 = 40.7 \, m$.

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