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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?
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The correct answer is:
$36.4 \mathrm{~km}$
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 $=s_1+s_2$ $=36367.3 \mathrm{~m}=36.4 \mathrm{~km}$
$s_1=u t+\frac{1}{2} a t^2=0+\frac{1}{2} \times 10 \times(60)^2=18000 \mathrm{~m}$
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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