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A ball is thrown straight up in air with an initial velocity u. Air exerts a force on it in horizontal direction which produces an acceleration depending on its height from ground as `a_(x) = ah^(2)`. Find the displacement of ball from the projection point as a function of time. |
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Answer» Correct Answer - [`r = sqrt(x_(t)^(2) + y_(t)^(2))` where `x_(t) = (au^(2)t^(4))/(12) + (ag^(2) t^(6))/(120) - (aug t^(5))/(20), y_(1) ut - (1)/(2) g t^(2)]` |
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