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Q.
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?
Height covered in 1 min,
$s_{1}=ut+\frac{1}{2}\,at^{2}=0+\frac{1}{2}\times10\times\left(60\right)^{2}=18000\,m$
Velocity attained after 1 min,
v = u + at = 0 + 10 × 60 = 600 m $s^{-1}$
After the fuel is finished, u = 600 m $^{s-1}$, v = 0
v$^{2} - u^{2 }= 2gs_{2}$
or$\quad 0 - \left(600\right)^2 = 2 × \left(- 9.8\right) × s_{2}$
or$\quad s_{2}=\frac{\left(600\right)^{2}}{2\times9.8}=18367.3\,m$
Maximum height reached = $s_{1} + s_{2}$
= 36367.3 m = 36.4 km