Physics - Kinematics Question with Solution | TestHub

Let$t$be the time interval. Then,



$2h=\left(u_x\right)\left(t\right)$

or$u_x=\frac{2h}{t}$...(i)

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

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

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

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

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

or$u_y^2=\frac{g^2{\left(t\right)}^2}{4}+2gh$...(ii)

Given,$u_x^2+u_y^2={\left(2\sqrt{gh}\right)}^2$

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

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

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

or$t=2\sqrt{\frac{h}{g}}$