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Question Number 134369 by I want to learn more last updated on 02/Mar/21

Answered by Dwaipayan Shikari last updated on 03/Mar/21

$$VerticalVelocity$$$$v^2=u^2+2gh\Rightarrow v=\sqrt{2gh}u=0$$$$v=\sqrt{16g}$$$$Netvelocity=\sqrt{{12}^2+v^2}=\sqrt{144+16g}=4\sqrt{g+9}m/s$$$$Angle={tan}^{-1}\frac{12}{4\sqrt{g}}={tan}^{-1}\frac{3}{\sqrt{g}}$$

Commented by I want to learn more last updated on 03/Mar/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{apprecate}.$$

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