A body is thrown vertically upwards from A, the top of the tower,...

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}$_{0} \mathrm{t}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2} \quad \ldots(1)$ 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} \quad \therefore \mathrm{t}=\sqrt{\mathrm{t}_{1} \mathrm{t}_{2}}$