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A rocket is fired vertically up from the ground with an acceleration of 10m/s2. If its fuel is finished after 1 minute then calculate-
(a) Maximum velocity attained by rocket in ascending motion.
(b) Height attained by rocket before fuel is finished.
(c) Time taken by the rocket in the whole motion.
(d) Maximum height attained by rocket.

Answer
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Hint: Firstly you could draw a rough diagram representing the whole motion. Then you could use Newton’s equations of motion accordingly to solve the given questions. Also remember that the initial velocity when fired as well as the velocity at maximum height is zero for the rocket.

Formula Used:
Newton’s equations of motion,
v=u+at
s=ut+12at2
v2u2=2as

Complete step by step answer:
In the question, we have a rocket that is fired vertically upward with an acceleration of 10ms2 and this rocket runs out of fuel after 1 minute. Even after completion of the fuel, we know that, due to inertia the rocket travels a short distance upward to reach a maximum height. So the situation can be represented as,
seo images

(a) Maximum velocity attained by rocket in ascending motion.
We know that initial velocity is zero when the rocket is fired. So,
u=0
From Newton’s equations of motion we have,
v=u+at ………………… (1)
Where, v is the final velocity, ‘u’ is the initial velocity, ‘a’ is the acceleration and t is the time taken.
We are told that the rocket runs out of fuel in 1minute and we have to find the maximum velocity attained during that time interval. So,
tY=1min=60s
Also, acceleration is given as,
a=10ms2
Now (1) becomes,
vy=0+(10)(60)=600ms1
(b) Height attained by rocket before fuel is finished
In the figure Y is point where the rocket runs out of fuel, so we are to find the distance from X to Y, that is, h1
From Newton’s equations,
s=ut+12at2 ……………………………. (2)
h1=0+12×10×(60)2
h1=18000m=18km
(d) Maximum height attained by rocket.
Let us findh2. At point Y the rocket has maximum velocity vy and at point Z, that is, at the maximum height the velocity is zero.
From Newton’s equations,
v2u2=2as
02vy2=2gh2
h2=(600ms1)22×10ms2=18000m=18km
Maximum height reached by rocket is the distance from X to Z,
Hmax=h1+h2=18km+18km
Hmax=36km
(c) Time taken by the rocket in the whole motion.
Time taken to travel from Y to Z can be given by,
From (1),
0=vygt
tYZ=vyg=60010=60s
Time taken for the return journey,
From (2),
Hmax=0+12gtr2
tr2=2Hmax10=2×3600010
tr=602s
The total time taken by rocket for whole motion T is,
T=tY+tyz+tr
T=(60+60+602)s
T=60(2+2)s
T=(2+2)min

So we have the solutions as,
(a) Maximum velocity attained by rocket in ascending motion is 600ms1
(b) Height attained by rocket before fuel is finished 18km
(c) Time taken by the rocket in the whole motion (2+2)min
(d) Maximum height attained by rocket 36km


Note:
After the fuel is used up, the acceleration of the rocket is that due to gravity (g). But we have to be careful about the direction as by convention we take g as positive downward. Hence, we have taken g as positive while calculating the time taken for return journey and we have taken g as negative while calculating h2 and tYZ.