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A stone is projected horizontally with a velocity 9.8ms^–1 from a tower of height 100m. Its velocity one second after projection is

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9.8 ms^-1

4.9 $\sqrt{2}$ ms^-1

4.9 ms^-1

9.8 $\sqrt{2}$ ms^-1

answer is C.

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Detailed Solution

$\begin{array}{l} V = \sqrt{(u)^{2} + (gt)^{2}} = \sqrt{(9.8)^{2} + (9.8)^{2}} \\ = 9.8 \sqrt{2} {ms}^{- 1} \end{array}$

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