Question Number 44730 by Necxx last updated on 03/Oct/18
Commented by Necxx last updated on 03/Oct/18
$${please}\:{help}\:{with}\:{no}.\:\mathrm{5} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Oct/18
![s=ut+(1/2)gt^2 distance in t second distance covered in t−1 second s_(t−1) =u(t−1)+(1/2)g(t−1)^2 s−s_(t−1) =distance covered in last one second =u(t−t+1)+(1/2)g(t^2 −t^2 +2t−1) s_(last) =u+(1/2)g(2t−1) now s=h s_(last) =((9h)/(25)) (s_(last) /s)=(((9h)/(25))/h)=(((1/2)g(2t−1))/((1/2)gt^2 )) [ u=0 ] (9/(25))=((2t−1)/t^2 ) 9t^2 −50t+25=0 t=((50±(√(2500−900)))/(18)) t=(((50+40)/(18))), (((50−40)/(18))) t=5 and {(5/9)←not feasible} h=(1/2)×9.8×5^2 =122.5](https://www.tinkutara.com/question/Q44736.png)
$${s}={ut}+\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \:{distance}\:{in}\:{t}\:{second} \\ $$$$ \\ $$$${distance}\:{covered}\:{in}\:{t}−\mathrm{1}\:{second}\: \\ $$$${s}_{{t}−\mathrm{1}} ={u}\left({t}−\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}{g}\left({t}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${s}−{s}_{{t}−\mathrm{1}} ={distance}\:{covered}\:{in}\:{last}\:{one}\:{second} \\ $$$$={u}\left({t}−{t}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}{g}\left({t}^{\mathrm{2}} −{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}\right) \\ $$$${s}_{{last}} ={u}+\frac{\mathrm{1}}{\mathrm{2}}{g}\left(\mathrm{2}{t}−\mathrm{1}\right) \\ $$$${now}\:{s}={h}\:\:\:\:{s}_{{last}} =\frac{\mathrm{9}{h}}{\mathrm{25}} \\ $$$$\frac{{s}_{{last}} }{{s}}=\frac{\frac{\mathrm{9}{h}}{\mathrm{25}}}{{h}}=\frac{\frac{\mathrm{1}}{\mathrm{2}}{g}\left(\mathrm{2}{t}−\mathrm{1}\right)}{\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\left[\:{u}=\mathrm{0}\:\:\right] \\ $$$$\frac{\mathrm{9}}{\mathrm{25}}=\frac{\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$$\mathrm{9}{t}^{\mathrm{2}} −\mathrm{50}{t}+\mathrm{25}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{50}\pm\sqrt{\mathrm{2500}−\mathrm{900}}}{\mathrm{18}} \\ $$$${t}=\left(\frac{\mathrm{50}+\mathrm{40}}{\mathrm{18}}\right),\:\left(\frac{\mathrm{50}−\mathrm{40}}{\mathrm{18}}\right) \\ $$$${t}=\mathrm{5}\:{and}\:\left\{\frac{\mathrm{5}}{\mathrm{9}}\leftarrow{not}\:{feasible}\right\} \\ $$$${h}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{9}.\mathrm{8}×\mathrm{5}^{\mathrm{2}} \:=\mathrm{122}.\mathrm{5} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Necxx last updated on 04/Oct/18
$${yeah}…{Thank}\:{you}\:{so}\:{much}. \\ $$