A rocket is fired vertically from the ground with a resultant vertical acceleration of a \(10…
A rocket is fired vertically from the ground with a resultant vertical acceleration of a \(10 \mathrm{~ms}^{-2}\). Fuel is finished in \(1 \mathrm{~min}\) and it continues to move up. What is the maximum height reached ?
\(36.4 \mathrm{~km}\)
\(42.3 \mathrm{~km}\)
\(48.4 \mathrm{~km}\)
\(25.6 \mathrm{~km}\)
Solution
Resultant vertical acceleration,
\(a=10 \mathrm{~ms}^{-2}\)
Height travelled by the rocket in \(1 \mathrm{~min}(60 \mathrm{~s})\),
\(\begin{aligned}
h & =u t+\frac{1}{2} a t^2 \\
& =0 \times 60+\frac{1}{2} \times 10 \times(60)^2 \\
& =18000 \mathrm{~m}=18 \mathrm{~km}
\end{aligned}\)
Velocity of rocket after \(1 \mathrm{~min}(60 \mathrm{~s})\),
\(v=u+a t=0+10 \times 60=600 \mathrm{~ms}^{-1}\)
After \(1 \mathrm{~min}\), when fuel of rocket is finished, then
\(u=600 \mathrm{~ms}^{-1}, a=-g=-9.8 \mathrm{~m} / \mathrm{s}^2\)
Hence, if rocket goes at height \(h_2\), then
\(\begin{aligned}
v^2-u^2 & =2 g h_2 \\
0-(600)^2 & =2(-9.8) h_2 \\
\Rightarrow \quad h_2 & =\frac{360000}{2 \times 9.8} \\
& =18367.3 \mathrm{~m}=18.4 \mathrm{~km}
\end{aligned}\)
\(\therefore\) Maximum height travelled by rocket from ground \(=h_1+h_2=18+18.4=36.4 \mathrm{~km}\)