TS EAMCET · Physics · Motion In Two Dimensions
An archer shoots an arrow from a height \(4.2 \mathrm{~m}\) above the ground with a speed \(40 \mathrm{~m} / \mathrm{s}\) and at an angle \(30^{\circ}\) as shown in the figure. Determine the horizontal distance \(R\) covered by the arrow, when it hits the ground, (Take \(g=10 \mathrm{~m} / \mathrm{s}^2\) ) 
- A \(\frac{185}{\sqrt{3}} m\)
- B \(84 \sqrt{3} \mathrm{~m}\)
- C \(68 \sqrt{3} \mathrm{~m}\)
- D \(\frac{95}{\sqrt{3}} \mathrm{~m}\)
Answer & Solution
Correct Answer
(B) \(84 \sqrt{3} \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
Given, speed of arrow, \(v=40 \mathrm{~m} / \mathrm{s}\) and \(\theta=30^{\circ}\) Horizontal range \(R_1\) covered by the arrow is given by…
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