Diwali rocket when ignited at the ground level rises vertically above the level of a window 10 from the ground find the magnitude of velocity of the rocket at the time of its take off.
Answers
Answer:
Let the rocket acquire a speed v at the time when all its fuel gets exhausted. Referring to the v-t graph in Fig., the slope of OP gives the acceleration a1.
The slope of OP gives the acceleration
a1=tanΘ1=t1v ...(i)
The slope of PQ gives the acceleration
−(a2)=tanΘ2=t2−v ...(ii)
From (i) and (ii), we have
v=a1t2=a2t2. ...(iii)
The area under v - t graph = Area of ΔOPQ = (1/2) OQ.PM = s
Substituting OQ = t1+t2 and PM = v,
The total displacement s =(v/2)(t1+t2 ...(iv)
Substituting t1=a1v and t2=a2v from(iii) in (iv), we have
s=2v(a1v+a2v)
Finally substituting s = +h (as the area is positive), a
Answer:
Explanation:
Since the body is moving up, the acceleration due to gravity 'g' is taken with a negative sign in the equation of motion. The body rises up to a height of 10m, i.e., its velocity at this height is zero, substituting these values in third equation of motion.
