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An astronaut is on the surface of other planet whose air resistance is negligible. To measure the acceleration due t o gravity `(g)`, he throws a stone upwards. He observer that the stone reaches to a maximum height of `10m` and reaches the surface `4` second after it was thrown. find the accelertion due to gravity `(g)` on the surface of that planet in `m//s^(2)`. |
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Answer» Correct Answer - `5` `(u^(2))/(2g) = 10 m` `(2u)/(g) = 4sec` Solving, `g = 5 m//s^(2)` |
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