200 Followers Problem!

The problem poser has unnecessarily complicated the problem.Here's an easier solution :

Consider the situation till the fuel of rocket is exhausted.Here ,

$u_1=0m/s \quad ,\quad a_1=20 m/s^2 \quad , \quad t_1=60 sec \\ \Rightarrow v_1 =u_1 + a_1t_1 \\ \Rightarrow v_1=0+(20)(60) \\ \Rightarrow v_1=1200m/s$

Consider the case when the rocket gets stationary at the highest point since the fuel is exhausted. Here :

$u_2=v_1=1200m/s \quad , \quad v_2=0m/s^2 \quad, \quad a_2=-10m/s^2\quad,\quad t_2=x \\ \Rightarrow v_2=u_2+a_2t_2 \\ \Rightarrow 0 = 1200+(-10)(x) \\ \Rightarrow -1200=-10x \\ \Rightarrow x=120sec=2mins \\ \Rightarrow \Large\boxed{200x=400}$

Since the Rocket Starts from the ground, therefore, u=0

t=60 Seconds.

Acceleration=20 ${ms}^{-2}$

According to the first Equation of Motion, $v=u+at$

Therefore, Speed of the rocket= 1200 m/s after the fuel is exhausted.

Now, After the fuel is exhausted, Initial Speed $u$ = 1200 m/s

Final Speed $v$ = 0 m/s.

Acceleration= $-10{ms}^{-2}$

According to the third equation of Motion,

${v}^{2}-{u}^{2}=2as$ a=acceleration

s= Distance.

= $0-{1200}^{2}$ = $2*-10*x$

x=72000 m

Now, Using the Third Equation of motion,

72000= 1200*t+ $\dfrac{1}{2}*-10*{t}^{2}$

72000= $1200t-5{t}^{2}$

= ${5t}^{2}-1200t+7200$

= ${t}^{2}-240t+14400=0$

{(t-120)}^{2}=0)

t= 120 seconds = 2 minutes.

Now, $200*2=400$ , Which is our answer.

Cheers!