Physics - Kinematics Question with Solution | TestHub

Let the body is projected vertically upwards from A with a speed $u_0$ Using equation, $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$ For case $(1)-\mathrm{h}={\mathrm{u}}_0{\mathrm{t}}_1-\frac{1}{2}{\mathrm{gt}}_1^2\dots (1)$ For case (2) $-\mathrm{h}={\mathrm{u}}_0{\mathrm{t}}_2-\frac{1}{2}{\mathrm{gt}}_1^2$ Subtracting eq (2) from (1), we get $$\begin{array}{l}0={\mathrm{u}}_0\left({\mathrm{t}}_2+{\mathrm{t}}_1\right)+\frac{1}{2}\mathrm{g}\left({\mathrm{t}}_2^2-{\mathrm{t}}_1^2\right)\\ \Rightarrow {\mathrm{u}}_0=\frac{1}{2}\mathrm{g}\left({\mathrm{t}}_1-{\mathrm{t}}_2\right)\end{array}$$ Putting the value of ${\mathrm{u}}_0$ in eq (2), we get $$-\mathrm{h}=-\left(\frac{1}{2}\right)\mathrm{g}\left({\mathrm{t}}_1-{\mathrm{t}}_2\right){\mathrm{t}}_2-\left(\frac{1}{2}\right){\mathrm{gt}}_2^2$$ $$\Rightarrow \mathrm{h}=\frac{1}{2}\mathrm{g}\left({\mathrm{t}}_1{\mathrm{t}}_2\right)$$ For case $3,{\mathrm{u}}_0=0,\mathrm{t}=?$ $$-\mathrm{h}=0\times \mathrm{t}-\left(\frac{1}{2}\right){\mathrm{gt}}^2$$ $\mathrm{h}=\left(\frac{1}{2}\right){\mathrm{gt}}^2$ Comparing eq. (4) and $(5),$ we get $$\frac{1}{2}{\mathrm{gt}}^2=\frac{1}{2}{\mathrm{gt}}_1{\mathrm{t}}_2\therefore \mathrm{t}=\sqrt{{\mathrm{t}}_1{\mathrm{t}}_2}$$