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A body is projected horizontally from a height of 78.4 m with a velocity $10 {ms}^{- 1}$ . Its velocity after 3 seconds is : $[g = 10 {ms}^{- 2}]$ (Take direction of projection on $\vec{i}$ and vertically upward direction on $\vec{j}$ ).

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

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

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

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

answer is A.

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Detailed Solution

h = 78.4 m, u = 10 m/s, t = 3s, g = 10 {m/s}^{2}

$\vec{v} = v_{x} \hat{i} + v_{y} \hat{j}$

$v_{x} = 10$

$v_{y} = g t = 10 \times 3 = 30$

$\vec{v} = v_{x} \hat{i} + v_{y} \hat{j}$

$\vec{v} = 10 \hat{i} - 30 \hat{j}$

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