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A body is projected horizontally from a height of 78.4m with a velocity 10m/s. Its velocity after 3 seconds is….

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$10 \hat{i} - 30 \hat{j}$

$10 \hat{i} + 30 \hat{j}$

$20 \hat{i} - 30 \hat{j}$

$10 \hat{i} + 10 \sqrt{3} \hat{j}$

answer is A.

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Detailed Solution

$V_{x} = U_{x} = 10, V_{y} = u + g t$

After 3 seconds, $g t = 30$

$\Rightarrow {\vec{V}}_{x} = {\vec{U}}_{x} = 10 \hat{i}, {\vec{V}}_{y} = 30 (- \hat{j})$

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