Question

A body takes T₁to fall on ground when projected vertically upwards from a tower with speed u and takes time T₂to fall down when thrown down the tower with speed same u.
find time taken for its free fall from tower in terms ofT_1andT₂

Answer 1

$$\begin{gathered} \mathrm{Suppose}\mathrm{h}\mathrm{is}\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{tower}. \\ (\mathrm{Taking}\mathrm{downward}\mathrm{direction}\mathrm{as}\mathrm{positive}) \\ \mathrm{For}\mathrm{first}\mathrm{case}, \\ \mathrm{h}=-{\mathrm{uT}}_1+\frac{1}{2}{{\mathrm{gT}}_1}^2----\left(\mathrm{i}\right) \\ \mathrm{For}\mathrm{second}\mathrm{case}, \\ \mathrm{h}={\mathrm{uT}}_2+\frac{1}{2}{{\mathrm{gT}}_2}^2-----\left(\mathrm{ii}\right) \\ \mathrm{For}\mathrm{free}\mathrm{fall}, \\ \mathrm{h}=\frac{1}{2}{\mathrm{gT}}^2---\left(\mathrm{iii}\right) \\ \mathrm{From}\mathrm{equations}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{iii}\right), \\ -{\mathrm{uT}}_1+\frac{1}{2}{{\mathrm{gT}}_1}^2=\frac{1}{2}{\mathrm{gT}}^2 \\ \mathrm{So},{\mathrm{uT}}_1=\frac{1}{2}\mathrm{g}[{{\mathrm{T}}_1}^2-{\mathrm{T}}^2]---\left(\mathrm{iv}\right) \\ \mathrm{Similarly}\mathrm{from}\mathrm{equations}\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right), \\ {\mathrm{uT}}_2+\frac{1}{2}{{\mathrm{gT}}_2}^2=\frac{1}{2}{\mathrm{gT}}^2 \\ \mathrm{Or}{\mathrm{uT}}_2=\frac{1}{2}\mathrm{g}[{\mathrm{T}}^2-{{\mathrm{T}}_2}^2]----\left(\mathrm{v}\right) \\ \mathrm{From}\left(\mathrm{iv}\right)\mathrm{and}\left(\mathrm{v}\right), \\ \frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}=\frac{[{{\mathrm{T}}_1}^2-{\mathrm{T}}^2]}{[{\mathrm{T}}^2-{{\mathrm{T}}_2}^2]} \\ \mathrm{or}{\mathrm{T}}_1[{\mathrm{T}}^2-{{\mathrm{T}}_2}^2]={\mathrm{T}}_2[{{\mathrm{T}}_1}^2-{\mathrm{T}}^2] \\ \mathrm{Or}\mathrm{T}=\sqrt{{\mathrm{T}}_1{\mathrm{T}}_2} \end{gathered}$$