Question

An elevator in which a man is standing is moving upwards with a speed of 10 m/s .If the man drops a coin from height of 2.45 m, it reaches the floor of the elevator after a time (g=9.8 m/s2)

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12
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2.

Answer 1

$$\begin{gathered} \mathrm{initial}\mathrm{velocity}\mathrm{of}\mathrm{coin}=\mathrm{u}=-10{\mathrm{ms}}^{-1} \\ \mathrm{height}=\mathrm{h}=2.45\mathrm{m} \\ \mathrm{therefore}, \\ 2.45=10\mathrm{t}+\frac{1}{2}\times 9.8\times {\mathrm{t}}^2 \\ \mathrm{or}490{\mathrm{t}}^2-1000\mathrm{t}-245=0 \\ \mathrm{or}98{\mathrm{t}}^2-200\mathrm{t}-49=0 \\ \mathrm{or}\mathrm{t}=\frac{200\pm \sqrt{40000+19208}}{196}=\frac{200\pm \sqrt{59208}}{196} \\ \mathrm{On}\mathrm{deleting}\mathrm{negative}\mathrm{sign}\mathrm{we}\mathrm{get}, \\ \mathrm{t}=2.26\mathrm{second} \end{gathered}$$