AP EAMCET · PHYSICS · Motion In Two Dimensions
If the horizontal range of a body projected with a velocity ' \(u\) ' is 3 times the maximum height reached by it, then the range of the body is (g - acceleration due to gravity)
- A \(\frac{2 \mathrm{u}^2}{3 \mathrm{~g}}\)
- B \(\frac{4 \mathrm{u}^2}{5 \mathrm{~g}}\)
- C \(\frac{12 \mathrm{u}^2}{13 \mathrm{~g}}\)
- D \(\frac{24 \mathrm{u}^2}{25 \mathrm{~g}}\)
Answer & Solution
Correct Answer
(D) \(\frac{24 \mathrm{u}^2}{25 \mathrm{~g}}\)
Step-by-step Solution
Detailed explanation
\(R = 3H\) \(\frac{u^2 \sin(2\theta)}{g} = 3 \frac{u^2 \sin^2(\theta)}{2g}\) \(2 \sin(\theta) \cos(\theta) = \frac{3}{2} \sin^2(\theta)\) \(\tan(\theta) = \frac{4}{3}\) \(\sin(\theta) = \frac{4}{5}, \cos(\theta) = \frac{3}{5}\) \(R = \frac{u^2 (2 \sin(\theta) \cos(\theta))}{g}\)…
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