Question

A rocket, when fired, rises up with an acceleration of $20m/s^2$ 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?
$(g=10m/s^2)$

Answer 1

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/$s^2$ away from the earth. The total distance travelled by the rocket during first 60 seconds will be :

$x_1=ut+\frac{1}{2}a{t}^2$

$=0\times 1.0+\frac{1}{2}\times 20\times {60}^2$

$=36000m$

After this it behaves as a freely falling body.So, the acceleration will be $10m/s^2$ towards the earth.

Now, at maximum height the velocity of the rocket will be $0m/s^2$.

$\therefore$ for $x_2$ :

we will first find the velocity of the object after 1.0 minute let's call it $v1$ .

$v_1=u+at$

$v_1=0+20\times 60$

$v_1=1200m/s$

now,

$\because v^2=u^2+2ax$

$v_2^2=v_1^2-2g{x}_2$

$0^2=(1200{)}^2-2\times 10x_2$

$\Rightarrow x_2=72000m$

So, maximum height will be $x_1+x_2$

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 $v=1200m/s.$.

$\because x=ut+\frac{1}{2}g{t}^2$

$-36=1200t-\frac{1}{2}\times 10t^2$

$\Rightarrow 5t^2-1200t-36=0$

solving the quadratic equation we get $t=240s$ and $t=-0.029s$ .

Since time can't be negative.

total time$=240s+60s.$

total time $=300s$.