TS EAMCET · Physics · Motion In Two Dimensions
Two bodies are projected from the same point with the same initial velocity ' \(u\) ' making angles ' \(\theta\) ' and \(\left(90^{\circ}-\theta\right)\) with the horizontal in opposite directions. The horizontal distance between their positions when the bodies are at their maximum heights is
- A \(\frac{\mathrm{u}^2}{2 \mathrm{~g}}\left(\sin ^2 \theta-\cos ^2 \theta\right)\)
- B \(\frac{\mathrm{u}^2 \sin 2 \theta}{2 \mathrm{~g}}\)
- C \(\frac{\mathrm{u}^2}{\mathrm{~g}}\)
- D \(\frac{\mathrm{u}^2 \sin 2\left(90^{\circ}-\theta\right)}{\mathrm{g}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{u}^2 \sin 2\left(90^{\circ}-\theta\right)}{\mathrm{g}}\)
Step-by-step Solution
Detailed explanation
Horizontal position at max height for angle \(\alpha\): \(x_h = \frac{u^2 \sin \alpha \cos \alpha}{g}\) For first body (\(\alpha = \theta\)): \(x_1 = \frac{u^2 \sin \theta \cos \theta}{g}\) For second body (\(\alpha = 90^{\circ}-\theta\)):…
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