Question

A stone falls freely from rest and the total distance covered by it in last second of it motion equals the distance covered by it in the first 3s of its motion. How long the the stone will remain in the air?

Answer 1

$$\begin{gathered} \mathrm{Answer} \\ \mathrm{Let}\mathrm{the}\mathrm{stone}\mathrm{falls}\mathrm{for}\mathrm{t}\mathrm{second} \\ \mathrm{Distance}\mathrm{fall}\mathrm{in}(\mathrm{t}-1)\mathrm{second}={\mathrm{s}}_{(\mathrm{t}-1)} \\ {\mathrm{s}}_{(\mathrm{t}-1)}=\mathrm{u}(\mathrm{t}-1)+\frac{1}{2}\mathrm{g}(\mathrm{t}-1{)}^2 \\ {\mathrm{s}}_{(\mathrm{t}-1)}=\frac{1}{2}\mathrm{g}({\mathrm{t}}^2-2\mathrm{t}+1)......(1)\left(\mathrm{since}\mathrm{initial}\mathrm{velocity}\mathrm{u}=0\right) \\ \mathrm{distance}\mathrm{fall}\mathrm{in}\mathrm{t}\mathrm{second},s_t=\frac{1}{2}{\mathrm{gt}}^2.............(2) \\ \mathrm{therefore}\mathrm{distance}\mathrm{fall}\mathrm{in}\mathrm{last}\mathrm{second} \\ \mathrm{eqn}(2)-\mathrm{eqn}(1) \\ =\frac{1}{2}\mathrm{g}(2\mathrm{t}-1)...........3 \\ \mathrm{now}\mathrm{distance}\mathrm{fall}\mathrm{in}\mathrm{first}3\mathrm{second} \\ =\frac{1}{2}\mathrm{g}(3{)}^2=\frac{9\mathrm{g}}{2} \\ \mathrm{and}\mathrm{distance}\mathrm{fall}\mathrm{in}\mathrm{last}3\mathrm{second}=\frac{1}{2}\mathrm{g}(2\mathrm{t}-1) \\ \frac{9\mathrm{g}}{2}=\frac{\mathrm{g}}{2}(2\mathrm{t}-1) \\ 2\mathrm{t}=10 \\ \mathrm{t}=5\mathrm{second} \end{gathered}$$