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
$${Vertical}\:{Velocity}\: \\ $$$${v}^{\mathrm{2}} ={u}^{\mathrm{2}} +\mathrm{2}{gh}\Rightarrow{v}=\sqrt{\mathrm{2}{gh}}\:\:\:\:\:{u}=\mathrm{0} \\ $$$${v}=\sqrt{\mathrm{16}{g}} \\ $$$${Net}\:{velocity}\:=\sqrt{\mathrm{12}^{\mathrm{2}} +{v}^{\mathrm{2}} }=\sqrt{\mathrm{144}+\mathrm{16}{g}}=\mathrm{4}\sqrt{{g}+\mathrm{9}}\:\:{m}/{s} \\ $$$${Angle}\:={tan}^{−\mathrm{1}} \frac{\mathrm{12}}{\mathrm{4}\sqrt{{g}}}={tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\:\sqrt{{g}}} \\ $$
Commented by I want to learn more last updated on 03/Mar/21
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{apprecate}. \\ $$