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The stones pass each other after 3 seconds. What is the height of the building |
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Answer» Let Y1(t) be position of dropped stone and Y2(t) be position of projected stone. Then: \(Y_1(t) = H - \frac12gt^2\) \(Y_2(t) = v_ot - \frac12gt^2\) \(Y_1(3) = Y_2(3)\implies H - \frac92g = 20 \frac ms\cdot 3s - \frac92g\) Therefore \(H = 60m\) |
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