Question Number 20411 by Tinkutara last updated on 26/Aug/17
$$\mathrm{A}\:\mathrm{stone}\:\mathrm{of}\:\mathrm{weight}\:{W}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{straight} \\ $$$$\mathrm{up}\:\mathrm{from}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{with}\:\mathrm{an}\:\mathrm{initial} \\ $$$$\mathrm{speed}\:{u}.\:\mathrm{if}\:\mathrm{a}\:\mathrm{drag}\:\mathrm{force}\:\mathrm{of}\:\mathrm{constant} \\ $$$$\mathrm{magnitude}\:{f}\:\mathrm{acts}\:\mathrm{on}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{through} \\ $$$$\mathrm{out}\:\mathrm{its}\:\mathrm{flight},\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{stone}\:\mathrm{just} \\ $$$$\mathrm{before}\:\mathrm{reaching}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{is} \\ $$
Answered by ajfour last updated on 26/Aug/17
$$\mathrm{0}={u}^{\mathrm{2}} −\mathrm{2}\left(\frac{{W}+{f}}{{W}}\right){gh}\:\:\left({for}\:{upward}\:{trip}\right) \\ $$$${v}^{\mathrm{2}} =\mathrm{2}\left(\frac{{W}−{f}}{{W}}\right){gh}\:\:\:\left({for}\:{downward}\:{trip}\right) \\ $$$$\Rightarrow\left(\frac{{W}−{f}}{{W}}\right)\left({u}^{\mathrm{2}} \right)\left(\frac{{W}}{{W}+{f}}\right) \\ $$$$\:\:\Rightarrow\:\:\boldsymbol{{v}}=\boldsymbol{{u}}\sqrt{\frac{\boldsymbol{{W}}−\boldsymbol{{f}}}{\boldsymbol{{W}}+\boldsymbol{{f}}}}\:. \\ $$
Commented by Tinkutara last updated on 26/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$