TS EAMCET · Physics · Motion In One Dimension
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 \(\left.g=10 \mathrm{~m} / \mathrm{s}^2\right)\)
- A \(100 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
- B \(150 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
- C \(50 \sqrt{6} \mathrm{~m} / \mathrm{s}\)
- D \(75 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(C) \(50 \sqrt{6} \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
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,…
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