A rocket is fired vertically from the ground with a resultant vertical acceleration of 10 ~m ~s^ -2 . The fuel is finished in 1 ~min and it continues to move up. What is the maximum height reached?

Correct answer: C. 36.4 ~km Height covered in 1 ~min , s_1=u t+(1)/(2) a t^2=0+(1)/(2) × 10 ×(60)^2=18000 ~m Velocity attained after 1 ~min , v=u+a t=0+10 × 60=600 ~m ~s^ -1 After the fuel is finished, u=600 ~m ~s^ -1 , v=0 aligned & v^2-u^2=2 g s_2 & or 0-(600)^2=2 ×(-9.8) × s_2 & or s_2=((600)^2)/(2 × 9.8)=18367.3 ~m aligned Maximum height reached =s_1+s_2=36367.3 ~m=36.4 ~km

AIIMS 201826 May

Exam: AIIMS
Year: 2018
Date: 26 May
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

AIIMS 2018 · 26 May · Physics · Center of Mass, Momentum and Collision

A rocket is fired vertically from the ground with a resultant...

A rocket is fired vertically from the ground with a resultant vertical acceleration of

10 \mathrm{~m} \mathrm{~s}^{-2}

. The fuel is finished in

1 \mathrm{~min}

and it continues to move up. What is the maximum height reached?

Options

A $42.3 \mathrm{~km}$
B $48.4 \mathrm{~km}$
C $36.4 \mathrm{~km}$
D $25.6 \mathrm{~km}$

Correct answer

C. $36.4 \mathrm{~km}$

Step-by-step solution

Height covered in

1 \mathrm{~min}

s_1=u t+\frac{1}{2} a t^2=0+\frac{1}{2} \times 10 \times(60)^2=18000 \mathrm{~m}

Velocity attained after

1 \mathrm{~min}

v=u+a t=0+10 \times 60=600 \mathrm{~m} \mathrm{~s}^{-1}

After the fuel is finished,

u=600 \mathrm{~m} \mathrm{~s}^{-1}, v=0

\begin{aligned} & v^2-u^2=2 g s_2 \\ & \text{ or } \quad 0-(600)^2=2 \times(-9.8) \times s_2 \\ & \text{ or } \quad s_2=\frac{(600)^2}{2 \times 9.8}=18367.3 \mathrm{~m}\end{aligned}

Maximum height reached

=s_1+s_2=36367.3 \mathrm{~m}=36.4 \mathrm{~km}$

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