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A particle is projected vertically up with speed of ${10 ms}^{- 1}$ . The time after which it will pass through the point 3.2 m above the point of projection $(g = {10 ms}^{- 2})$

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0.4 s

0.5 s

1.2 s

1.8 s

answer is A.

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Detailed Solution

$h = ut - \frac{1}{2} g t^{2} 3.2 = 1 0t - {5t}^{2} {5t}^{2} - 10 t + 3.2 = 0 t = 1.6, 0.4 s$

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