A stone is dropped into water from a bridge 441 \(\pi\) above the water. Another stone is thrown vertically downward 1 sec later. Both strike the water simultaneously. What was the initial speed of the second stone
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Time taken by first stone to reach the water surface from the bridge be t , then
\(h - vt + \frac{1}{2}gt^{2} = 44.1 - 0t + \frac{1}{2}9.8t^{2}\)
\(\tau = \frac{12/44.1}{9.8} \quad 3 \ sec\)
Second stone is thrown 1 sec later and both strikes simultaneously. This means that the time left for second stone \(3 - 1 \quad 2 \ sec\)
Hence \(44.1 - v \times 2 + \frac{1}{2} 9.8 (2)^2\)
\(\rightarrow 44.1 - 19.6 \quad 2u \rightarrow u \quad 12.25 \, m/s\)
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