JEE Mains · Physics · STD 11 - 3.2 motion in plane
Two projectiles are fired from ground with same initial speeds from same point at angles \(\left(45^{\circ}+\alpha\right)\) and \(\left(45^{\circ}-\alpha\right)\) with horizontal direction. The ratio of their times of flights is
- A 1
- B \(\frac{1-\tan \alpha}{1+\tan \alpha}\)
- C \(\frac{1+\sin 2 \alpha}{1-\sin 2 \alpha}\)
- D \(\frac{1+\tan \alpha}{1-\tan \alpha}\)
Answer & Solution
Correct Answer
(D) \(\frac{1+\tan \alpha}{1-\tan \alpha}\)
Step-by-step Solution
Detailed explanation
\(\theta_1=45+\alpha ; \theta_2=45-\alpha\) Time of flight, \(\mathrm{T}=\frac{2 \mathrm{v} \sin \theta}{\mathrm{g}}\) \(\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\sin (45+\alpha)}{\sin (45-\alpha)}\)…
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