A rocket is fired upwards vertically with a net acceleration of 4m/s""^2 and initial velocity zero. After 5 seconds its fuel is finished and it decelerates with g. At the highest point its velocity becomes zero. There after it accelerates downwards with acceleration g and return back to ground. i) Plot velociry - time graph for complete journey ii) Displacement-time graph for the complete journey: ("Take g =10 m/s"^2)

A rocket is fired upwards vertically with a net acceleration of 4m/s""

1.	A rocket is fired upwards vertically with a net acceleration of 4m/s""^2 and initial velocity zero. After 5 seconds its fuel is finished and it decelerates with g. At the highest point its velocity becomes zero. There after it accelerates downwards with acceleration g and return back to ground. i) Plot velociry - time graph for complete journey ii) Displacement-time graph for the complete journey: ("Take g =10 m/s"^2)
Answer» Solution :Stage 1 : To find velocity of rocket after 5 seconds `V_(A)=0+at_(OA) =(4)(5)=20ms^(-1)` Stage II: To find further time of ascent after 5 seconds. `0=20-gt_(AB)""t_(AB)=(20)/(10)=2"seconds"` Here, the total VERTICAL displacement of stage (i) and stage (ii) is `="area of OAB"=1/2 (7) (20)=70m` Stage -III: If `t_(BC)` is time TAKEN of DESCENT then `70=1/2 (10) t_(BC)^(2) ` `t_(BC)=sqrt(14)-3.7s`