Physics - Kinematics Question with Solution | TestHub

$2h=ut$

$u_x=\frac{2h}{\mathrm{\Delta }t} \left[\because u=u_x,t=\mathrm{\Delta }t\right]....\left(\mathrm{i}\right)$

$h=u_{y}t-\frac{1}{2}g{t}^2$

$g{t}^2-2u_{y}t+2h=0$

$t_1=\frac{2u_y+\sqrt{4u_y^2-8gh}}{2g}$

$t_2=\frac{2u_y-\sqrt{4u_y^2-8gh}}{2g}$

$\mathrm{\Delta }t=t_1-t_2=\frac{\sqrt{4u_y^2-8gh}}{g}$

$u_y^2=\frac{g^2(\Delta t{)}^2}{4}+2gh$

$u_x^2+u_y^2=u^2=(2\sqrt{gh}{)}^2$

$\frac{4n^2}{(\mathrm{\Delta }t{)}^2}+\frac{g^2(\mathrm{\Delta }t{)}^2}{4}+2gh=4gh$

$\frac{g^2}{4}{\left(\mathrm{\Delta }t\right)}^4-2gh{\left(\mathrm{\Delta }t\right)}^2+4h^2=0$

$\left(\mathrm{\Delta }t\right)=\frac{2gh\pm \sqrt{4g^2{h}^2-4g^2{h}^2}}{g^2/2}=\frac{4h}{g}$

$\mathrm{\Delta }t=2\sqrt{\frac{h}{g}}$