Question

Two balls are projected simultaneously with the same velocity from the top of the tower one vertically upwords & other vertically downwords. If they reach the ground in times t1& t2 respectively, show that Root over ( t1t2) is the time which each will take to reach the ground if simply let drop from the top of the tower.

Answer 1

$$\begin{gathered} \mathrm{Suppose}\mathrm{h}\mathrm{is}\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{tower}. \\ \mathrm{h}=\frac{1}{2}{{\mathrm{gt}}_1}^2[\mathrm{Free}\mathrm{fall}] \\ \mathrm{h}={\mathrm{ut}}_2-\frac{1}{2}{{\mathrm{gt}}_2}^2[\mathrm{Thrown}\mathrm{downwards}] \\ {\mathrm{t}}_1={\mathrm{t}}_2+\frac{2\mathrm{u}}{\mathrm{g}} \\ \mathrm{or}\mathrm{u}=\frac{\mathrm{g}}{2}({\mathrm{t}}_1-{\mathrm{t}}_2) \\ \frac{1}{2}{\mathrm{gt}}^2={\mathrm{ut}}_2-\frac{1}{2}{{\mathrm{gt}}_2}^2 \\ \mathrm{or}\frac{1}{2}{\mathrm{gt}}^2=\frac{\mathrm{g}}{2}({\mathrm{t}}_1-{\mathrm{t}}_2){\mathrm{t}}_2-\frac{1}{2}{{\mathrm{gt}}_2}^2=\frac{{\mathrm{gt}}_1{\mathrm{t}}_2}{2} \\ \mathrm{or}\mathrm{t}=\sqrt{{\mathrm{t}}_1{\mathrm{t}}_2} \end{gathered}$$