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A balloon in ascending vertically with a...

A balloon in ascending vertically with an acceleration of `1 ms^(-2)`. Two stones are dropped from it at an interval of `2 s`. Find the distance berween them `1.5 s` after the second stone is released.

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To solve the problem, we need to analyze the motion of the two stones dropped from the ascending balloon. Let's break it down step by step. ### Step 1: Understand the motion of the balloon The balloon is ascending with an acceleration of \(1 \, \text{m/s}^2\). This means that at the moment the stones are dropped, they have the same upward velocity as the balloon. ### Step 2: Determine the time intervals The first stone is dropped at \(t = 0 \, \text{s}\), and the second stone is dropped at \(t = 2 \, \text{s}\). We need to find the distance between the two stones \(1.5 \, \text{s}\) after the second stone is released, which means we will analyze the motion of both stones at \(t = 3.5 \, \text{s}\) (since \(2 \, \text{s} + 1.5 \, \text{s} = 3.5 \, \text{s}\)). ...
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