Question Number 134369 by I want to learn more last updated on 02/Mar/21
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Answered by Dwaipayan Shikari last updated on 03/Mar/21
![Vertical Velocity v^2 =u^2 +2gh⇒v=(√(2gh)) u=0 v=(√(16g)) Net velocity =(√(12^2 +v^2 ))=(√(144+16g))=4(√(g+9)) m/s Angle =tan^(−1) ((12)/(4(√g)))=tan^(−1) (3/( (√g)))](https://www.tinkutara.com/question/Q134372.png)
$${Vertical}\:{Velocity}\: \\ $$$${v}^{\mathrm{2}} ={u}^{\mathrm{2}} +\mathrm{2}{gh}\Rightarrow{v}=\sqrt{\mathrm{2}{gh}}\:\:\:\:\:{u}=\mathrm{0} \\ $$$${v}=\sqrt{\mathrm{16}{g}} \\ $$$${Net}\:{velocity}\:=\sqrt{\mathrm{12}^{\mathrm{2}} +{v}^{\mathrm{2}} }=\sqrt{\mathrm{144}+\mathrm{16}{g}}=\mathrm{4}\sqrt{{g}+\mathrm{9}}\:\:{m}/{s} \\ $$$${Angle}\:={tan}^{−\mathrm{1}} \frac{\mathrm{12}}{\mathrm{4}\sqrt{{g}}}={tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\:\sqrt{{g}}} \\ $$
Commented by I want to learn more last updated on 03/Mar/21
![Thanks sir. I apprecate.](https://www.tinkutara.com/question/Q134394.png)
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{apprecate}. \\ $$