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A particle is projected vertically upwards from point A with initial speed of 29.4 m/s. During what interval of time is the particle at more than 39.2m from A. |
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Answer» Solution :`(u^2 sin^2 theta)//(2G)` = (u^2sin ^2THETA)//(2g)` gives `tantheta=4 or `theta= tan^-1(4)` |
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