A rocket is fired upwards vertically with a net acceleration of `4"m/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 velocity – time graph for complete journey
(ii) Displacement-time graph for the complete journey. (Take `g = 10 m//s^2`.)

Explanation

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`