A diwali rocket moves vertically with constant acceleration of 5 ms–2. After some time its fuel exhausted and then it falls freely. If maximum height attained by the rocket is 60m. Then find its speeds when the fuel is just exhausted. (g = 10ms–2)

Asked by aditvk27 | 21st Aug, 2021, 09:56: AM

Expert Answer:

Let h be the vertical distance travelled by rocket before exhausting the fuel.
 
Speed v of rocket after moved by a vertical distance h is determined from the following equation of motion
 
" v = u2 + ( 2 a S )  "  ...........................(1)
 
where u = 0 is initial speed because rocket starts from rest, a = 5 m/s2 is acceleration
and S = h is distance travelled.
 
v2 = 2 × 5 × h     or    v = begin mathsize 12px style square root of 10 space h end root end style  m/s
 
If total vertical distance travelled by rocket is 60 m , distance travelled vertically in gravitational field is (60-h) m .
 
When the rocket reaches maximum height , its final speed v = 0 . Hence in equation of motion (1) ,
 
if we substitute v = 0 and initial speed after exhausting fuel   u  = begin mathsize 12px style square root of 10 space h end root end style  m/s , acceleration a = -g = -10 m/s2 
and S = (60-h) m , then we get
 
10 h = 2 × 10 × (60-h)
 
we get h = 40 m from above expression
 
hence speed of rocket immediately after exhausting fuel , v = begin mathsize 12px style square root of 10 cross times 40 end root end style = 20 m/s 

Answered by Thiyagarajan K | 21st Aug, 2021, 01:14: PM