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Question

A rocket, when fired, rises up with an acceleration of 20m/s2 for the first 1.0 minute. After this its fuel is finished and it goes on rising up just like a free body. Upto what maximum height will it rise? How much total time will it take before falling on earth?
(g=10m/s2)

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Solution

The first question we have been asked is the maximum height of the rocket, so let's find it first.
We consider the direction towards the earth as +ve and away from earth be ve.Since, during the first 1.0 minute the acceleration is 20m/s2 away from the earth. The total distance travelled by the rocket during first 60 seconds will be :
x1=ut+12at2
=0×1.0+12×20×602
=36000m
After this it behaves as a freely falling body.So, the acceleration will be 10m/s2 towards the earth.
Now, at maximum height the velocity of the rocket will be 0m/s2.
for x2 :
we will first find the velocity of the object after 1.0 minute let's call it v1 .
v1=u+at
v1=0+20×60
v1=1200m/s
now,
v2=u2+2ax
v22=v212gx2
02=(1200)22×10x2
x2=72000m
So, maximum height will be x1+x2
maximum height = 108km.
Now the second part , the total time taken by the rocket.
We know that it spends 1.0 minute after takeoff, we have to find the time after this moment and the moment it touches the earth again,so let's call it t .
The toal displacement of the rocket after 1.0 minute is 32 km towards the earth. Let's call it x.Initial velocity will be v=1200m/s..
x=ut+12gt2
36=1200t12×10t2
5t21200t36=0
solving the quadratic equation we get t=240s and t=0.029s .
Since time can't be negative.
total time=240s+60s.
total time =300s.

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