question_answer

A toy rocket leaves the ground with zero initial velocity and accelerates for 5s at\[\text{2 m}/{{\text{s}}^{\text{2}}}\]. The fuel is finished at this moment and it continues to move up till its velocity becomes zero. Find the total time for which it is in the air before returning to the ground.

Answer:

\[\because \]velocity after\[5s=0+2\times 5=10m/s\] height upto this point \[=\frac{1}{2}a{{t}^{2}}=\frac{1}{2}\times 2\times {{(5)}^{2}}=25m\] further height travelled by the body after \[5s=\frac{{{0}^{2}}-{{10}^{2}}}{-2g}=5m\] \[\therefore \] Total height\[=25+5=30m\] Total time = time taken from A to B + time taken from B to max ht (C) and then to the ground. \[t=5+{{t}^{1}}.\]How to get  \[{{t}^{1}}\]? \[s=+25\text{ }m,\] \[t=10m/s,\,\,t={{t}^{1}}=?,\,\,g=10m/{{s}^{2}}.\] \[25=10{{t}^{1}}-\frac{1}{2}\times 10\times {{t}^{{{I}^{2}}}}\] \[\Rightarrow 25=10{{t}^{1}}-5{{t}^{1}}\Rightarrow 5=2{{t}^{1}}-{{t}^{{{I}^{2}}}}\] On solving this we get, \[{{t}^{1}}=1+\sqrt{6}s\] \[\therefore \]total time \[t=5+1+\sqrt{6}=6+\sqrt{6}s\]