question_answer

A stone thrown vertically upwards attains a maximum height of 45 m. In what time does the velocity of stone become half the velocity of its throw?                     (\[\text{g}=\text{1}0\,\text{m}{{\text{s}}^{-\text{2}}}\])

Answer:

Let the initial velocity be u. \[h=45m\] Initial velocity = u Final velocity\[=u/2\] Acceleration\[(a)=-10\,\,m/{{s}^{2}}\] \[t=?\] We know that, \[v=u=at\] \[\Rightarrow \frac{u}{2}=u-10t\Rightarrow 10t=\frac{u}{2}(or)\] \[t=\frac{u}{20}\]                                             ??. (1) How to get u? We know, \[h=\frac{{{u}^{2}}}{2g}\Rightarrow 45=\frac{{{u}^{2}}}{20}\] \[\Rightarrow u=\sqrt{2\times 10\times 45}=30m/s\]                     ?..(2) From (1) and (2), we have\[t=1.5s\]