question_answer 73)

                  A person in all elevator accelerating upwards with all acceleration of \[2\,\,m{{s}^{-2}}\], tosses a coin vertically upwards with a speed of \[20\,\,m{{s}^{-1}}\]. After how much time will the coin fall back into his hand ? \[(g=10\,\,m{{s}^{-2}})\]

Answer:

                  Net upward acceleration, \[a'=g+a=12\text{ }m{{s}^{1}}\]                 Time taken to reach highest point,                 \[t=\,\frac{\upsilon -u}{-a'}=\,\frac{0-20}{-12}\,=\frac{5}{3}\,s\]                 \ Time of ascent \[=\frac{5}{3}s\]                 Time after which coin will fall back into his                 hands \[=\,\frac{5}{3}+\frac{5}{3}=\frac{10}{3}=3.33s\]