question_answer

A healthy young man standing at a distance of 7m from a 11.8 m high building sees a kid slipping from the top floor. With what speed (assumed uniform) should he run to catch the kid at the arms height (1.8 m)?

Answer:

Let\[{{t}_{1}}\]in the time taken for the kid to reach the hands of a man and \[{{t}_{1}}=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times (11.8-1.8)}{9.8}}\] \[=\sqrt{\frac{20}{9.8}}=\sqrt{\frac{10}{4.9}}=\sqrt{\frac{100}{49}}\] \[{{t}_{1}}=\frac{10}{7}\] ??????..(1) let\[\Rightarrow \]in the time taken for the man to catch the body with a velocity u \[{{t}_{2}}=\frac{S}{u}=\frac{7}{u}\] ??????..(2) As per the condition, both \[=19.6\times 12(13)=24\times 19.6=470.4\text{ }m.\]are same hence, \[{{t}_{1}}={{t}_{1}}\Rightarrow \frac{10}{7}=\frac{7}{u}\,\,u=\frac{49}{10}=4.9\,m/s\]