Question

A body let fall from the top of a tower falls through 3/8 of its height during the last second of its fall.what is the height of tower

Answer 1

$$\begin{gathered} \mathrm{height}=\mathrm{h},\mathrm{time}=\mathrm{t} \\ \mathrm{u}=0 \\ \mathrm{case}1:\mathrm{top}\mathrm{to}\mathrm{ground}\mathrm{in}\mathrm{time}\mathrm{t}, \\ \mathrm{S}=\mathrm{u}+\frac{1}{2}{\mathrm{gt}}^2 \\ \mathrm{h}=0+\frac{1}{2}{\mathrm{gt}}^2=\frac{1}{2}{\mathrm{gt}}^2...............\left(1\right) \\ \mathrm{case}2:\mathrm{top}\mathrm{to}\mathrm{a}\mathrm{point}\mathrm{where}\mathrm{time}\mathrm{was}(\mathrm{t}-1)\sec \\ \mathrm{S}=\mathrm{u}+\frac{1}{2}{\mathrm{gt}}^2 \\ \mathrm{h}-\frac{3}{8}\mathrm{h}=0+\frac{1}{2}\mathrm{g}{\left(\mathrm{t}-1\right)}^2 \\ \frac{5}{8}\mathrm{h}=\frac{1}{2}\mathrm{g}{\left(\mathrm{t}-1\right)}^2 \\ \mathrm{h}=\frac{8}{10}\mathrm{g}{\left(\mathrm{t}-1\right)}^2...............\left(2\right) \\ \mathrm{from}\left(1\right)\mathrm{and}\left(2\right), \\ \frac{1}{2}{\mathrm{gt}}^2=\frac{8}{10}\mathrm{g}{\left(\mathrm{t}-1\right)}^2 \\ {\mathrm{t}}^2=\frac{8}{5}\left({\mathrm{t}}^2+1-2\mathrm{t}\right) \\ \frac{3}{8}{\mathrm{t}}^2-2\mathrm{t}+1=0 \\ 3{\mathrm{t}}^2-16\mathrm{t}+8=0 \\ \mathrm{Let}\mathrm{a}=3,\mathrm{b}=-16,\mathrm{c}=8 \\ \mathrm{D}={\mathrm{b}}^2-4\mathrm{ac} \\ \mathrm{D}={\left(-16\right)}^2-4\times 3\times 8 \\ \mathrm{D}=256-96 \\ \mathrm{D}=160 \\ \mathrm{t}=\frac{-\mathrm{b}\pm \sqrt{\mathrm{D}}}{2\mathrm{a}} \\ \mathrm{t}=\frac{16\pm 12.6}{6} \\ \mathrm{t}=4.76\sec \mathrm{or}\mathrm{t}=0.56\sec \\ \mathrm{After}\mathrm{substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{t}\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}, \\ \mathrm{h}=113.2\mathrm{m}\mathrm{or}\mathrm{h}=1.56\mathrm{m} \end{gathered}$$