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A rocket moves straight upward with zero initial velocity and with an acceleration $20 \mathrm{~m} / \mathrm{s}^2$. It runs out of fuel and stops accelerating at the end of $5^{\text {th }} \mathrm{sec}$. It reaches a maximum height and falls back to the earth. The speed when it hits the ground is.
$\left(\right.$ Take $\left.g=10 \mathrm{~m} / \mathrm{s}^2\right)$
Options:
$\left(\right.$ Take $\left.g=10 \mathrm{~m} / \mathrm{s}^2\right)$
Solution:
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Verified Answer
The correct answer is:
$50 \sqrt{6} \mathrm{~m} / \mathrm{s}$
We have
$$
\begin{aligned}
& \mathrm{V}_{\max }=0+20 \times 5=100 \mathrm{~m} / \mathrm{s} \\
& \text { and, }\left.\mathrm{S}\right|_{\mathrm{t}=5}=\frac{1}{2} \times 20 \times 5^2=250 \mathrm{~m}
\end{aligned}
$$
Speed with it will strike ground,
$$
\begin{array}{lll}
\mathrm{V}^2=100^2+2 \times 10 \times 250 & \\
\Rightarrow \quad V^2=10000+5000 \quad & \Rightarrow V^2=15000 \\
\Rightarrow \quad V^2=2500 \times 6 \quad \Rightarrow V=50 \sqrt{6} \mathrm{~m} / \mathrm{s}
\end{array}
$$
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