• # question_answer A ball is projected horizontally from the top  of a tower with a velocity of $4\text{ }m{{s}^{-1}}.$ The velocity of I the ball after$0.7\text{ }s\,(g=10\,m{{s}^{-2}})$                  A) $11\text{ }m{{s}^{-1}}$                B) $~10\text{ }m{{s}^{-1}}$C) $~8\text{ }m{{s}^{-1}}$             D) $~3\text{ }m{{s}^{-1}}$

Correct Answer: C

Solution :

Velocity of the ball after time t $=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{{{(4)}^{2}}+{{(gt)}^{2}}}$ $=\sqrt{16+{{(10\times 0.7)}^{2}}}=8m{{s}^{-1}}$ Hence, the correction option is [c].

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