Physics - Kinematics Question with Solution | TestHub

From A to B motion is uniform

so

$\mathrm{h}={\mathrm{Vt}}_1$..........(1)

Time intend from A to B From $B$ to $C$ motion of uniform acceleration taking ' $B$ ' on origin upward direction positive $\mathrm{y}={\mathrm{U}}_{\mathrm{y}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{y}}{\mathrm{t}}^2$

After time ${\mathrm{t}}_2$ from point B

$\mathrm{y}=-\mathrm{h}$

$-\mathrm{h}={\mathrm{Vt}}_2-\frac{1}{2}{\mathrm{gt}}_2^2$ ............(2)

For that total time

$\mathrm{t}={\mathrm{t}}_1+{\mathrm{t}}_2$.........(3)

By (1) \& (2) we get

$-{\mathrm{Vt}}_1={\mathrm{Vt}}_2-\frac{1}{2}{\mathrm{gt}}_2^2\Rightarrow \frac{1}{2}{\mathrm{gt}}_2^2=\mathrm{V}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)$

${\mathrm{t}}_2=\sqrt{\frac{2\mathrm{Vt}}{\mathrm{g}}}$ $\dots (4)\left({\mathrm{Ast}}_1+{\mathrm{t}}_2=\mathrm{t}\right)$

Now using (1) (3) \& (4) we get $\mathrm{t}=\frac{\mathrm{h}}{\mathrm{V}}+\sqrt{\frac{2\mathrm{Vt}}{\mathrm{g}}}$ $\mathrm{h}=\mathrm{v}\left[\mathrm{t}-\sqrt{\frac{2\mathrm{Vt}}{\mathrm{g}}}\right]=3$