A rocket is fired vertically upwards and Moves with a net acceleration of $20{ms}^{-2}$ After one minute, The fuel is exhausted. The time taken by the rocket to reach the highest point after the fuel is exhausted is nearly $x\ minutes$ .

Calculate $200*x$

Details and Assupmtions

Assume Gravity to be $10{ms}^{-2}$

The answer is 400.

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You and and also the Challenge Master are assuming
**
g
**
to be constant. But it reality,
*
for a rocket, especially
*
we have to take in account that
**
g
**
varies ( decreases ) as it gains height.

So NASA would
**
make a huge mess
**
if they use our this such easy calculations. They would need a little bit
**
Calculus
**
for this!!!

Muhammad Arifur Rahman
- 6 years ago

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Yeah you are certainly right.Don't take this problem with respect to real life.In real life , we must also have air resistance as the major factor.Yeah , the value of " $g$ " will decrease but this problem has been framed for constant acceleration.

If you are comparing this problem with the working of NASA,this comparison is absolutely inappropriate.There are lot of factors that NASA has to consider before experimentation.This problem is just a special case and has been framed for the understanding of physics learners.I hope you understand.Thanks!

Nihar Mahajan
- 6 years ago

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Ok. Then let's approach a bit further. You can make it more realistic, You just have to use this equation, $W=\text{GMm}(\frac{1}{a}-\frac{1}{b})$ .

Calculus would help.

I'm inspired from your brilliant problem and I'll be sharing a new problem with some more sophisticated calculations , and more realistically .

....Hope you enjoy solving that
**
Upcoming Problem!
**

Muhammad Arifur Rahman
- 6 years ago

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@Muhammad Arifur Rahman – Nice. I am eagerly waiting for it ☺

Nihar Mahajan
- 6 years ago

First, solve for initial velocity. v=at v=20(60)=1200m/s Then solve for distance the rocket goes with velocity initial of 1200 m/s. v=(2gx)^0.5 x=73469.3877 m Then use X=vt+0.5at^2 73469.3877=1200(t)+0.5(-9.8)(x^2) t=122.448 s This is around 2 minutes. 2 times 200 is 400 400 is the final answer.

Note I used g as 9.8 instead of 10.

William G.
- 5 years, 8 months ago

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!

1 Helpful
0 Interesting
0 Brilliant
0 Confused

**
Challenge student note:
**
Correct , but there was no need to compute the distance.Also , don't put
$"="$
sign before equations.

Nihar Mahajan
- 6 years, 1 month ago

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Okay! I'll keep that in mind the next time I post a solution! But, I'm too lazy to change it Right now :3 :3

Mehul Arora
- 6 years, 1 month ago

Congrats @Mehul Arora !

Sorry for commenting lately! I'd left your problem untouched, sorry for that. . . .

Please accept this:

$\huge CHEERS!!!$ $\huge CHEERS!!!$ $\huge CHEERS!!!$

Keep calm and reach $\infty$ followers!!!

Also keep calm and forgive me for responding late!!!

$\huge \ddot \smile$

Sravanth C.
- 6 years ago

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Ohh. Never mind, mate! and Thanks for all your good wishes!

Mehul Arora
- 6 years ago

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Welcome!

BTW, the question was tricky, which is why I left it, finally I solved at the $11^{th}$ hour!

Sravanth C.
- 6 years ago

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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 :

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