Saved Bookmarks
| 1. |
A rocket I fired upwards vertically with a net acceleration of 4m//s^(2) and initial velocity zero.After 5 seconds its fuel is finsihed and it deccelerated 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=10m//s^(2)) . |
|
Answer» Solution :STAGE:1To find velocity of ROCKET after 5 seconds. `V_(A)=0+at_(OA)=(4)(5)=20MS^(-1)` StageII:To find further TIME of ascent after 5 seconds . 0=2 `"gt"_(AB)""therefore t_(AB)=(20)/(10)=2seconds` Here,the total vertical displacement of stage (i) and Stage (ii) is=area of OAB `=(1)/(2)(7)(20)=70 m.` 
|
|