TS EAMCET · Physics · Motion In Two Dimensions
A bullet is fired at time \(t=0\) with velocity \(20 \mathrm{~m} / \mathrm{s}\) and at an initial angle of \(30^{\circ}\) with the horizontal. The an angle between the displacement vector and the horizontal after time \(0.1 \mathrm{~s}\) is (Assume \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(\frac{38}{20 \sqrt{3}}\)
- B \(\frac{19}{20 \sqrt{3}}\)
- C \(\frac{19}{20}\)
- D \(\frac{19 \sqrt{3}}{20}\)
Answer & Solution
Correct Answer
(B) \(\frac{19}{20 \sqrt{3}}\)
Step-by-step Solution
Detailed explanation
\(x=20 \cos 30^{\circ} \times 0.1=2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3} \mathrm{~m}\) \[ \begin{aligned} & \text { and, } y=20 \sin 30^{\circ} \times 0.1-\frac{1}{2} \mathrm{~g}(0.1)^2 \\ & =1-0.05=0.95 \mathrm{~m}=\frac{19}{20} \mathrm{~m} \end{aligned} \] So,…
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