question_answer

A body, dropped from a height h at time t = 0, reaches the ground at time\[{{t}_{0}}.\] At what time will it reach a height\[\frac{h}{2}\]? (g is assumed to be constant)

Answer:

We know that\[h=\frac{1}{2}g{{t}^{2}}\Rightarrow h\propto {{t}^{2}}\] (\[\because \]g assumed to be constant) \[\therefore \frac{{{h}_{1}}}{{{g}_{2}}}=\frac{t_{1}^{2}}{t_{1}^{2}}\Rightarrow \frac{h}{\frac{h}{2}}\Rightarrow {{t}^{2}}=\frac{t_{0}^{2}}{2}\therefore t=\frac{{{t}_{0}}}{\sqrt{2}}\]