Q65 / 75 · JEE Main 2025 (7 Apr Shift 1) · Physics
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
- A1
- B$\frac{1-\tan \alpha}{1+\tan \alpha}$
- C$\frac{1+\sin 2 \alpha}{1-\sin 2 \alpha}$
- Correct answerD$\frac{1+\tan \alpha}{1-\tan \alpha}$
Solution
$\theta_1=45+\alpha ; \theta_2=45-\alpha$
Time of flight, $\mathrm{T}=\frac{2 \mathrm{v} \sin \theta}{\mathrm{g}}$
$\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\sin (45+\alpha)}{\sin (45-\alpha)}$
$\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\frac{1}{\sqrt{2}} \cos \alpha+\frac{1}{\sqrt{2}} \sin \alpha}{\frac{1}{\sqrt{2}} \cos \alpha-\frac{1}{\sqrt{2}} \sin \alpha}$
$\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\cos \alpha+\sin \alpha}{\cos \alpha-\sin \alpha}=\frac{1+\tan \alpha}{1-\tan \alpha}$
Time of flight, $\mathrm{T}=\frac{2 \mathrm{v} \sin \theta}{\mathrm{g}}$
$\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\sin (45+\alpha)}{\sin (45-\alpha)}$
$\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\frac{1}{\sqrt{2}} \cos \alpha+\frac{1}{\sqrt{2}} \sin \alpha}{\frac{1}{\sqrt{2}} \cos \alpha-\frac{1}{\sqrt{2}} \sin \alpha}$
$\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\cos \alpha+\sin \alpha}{\cos \alpha-\sin \alpha}=\frac{1+\tan \alpha}{1-\tan \alpha}$
Hence, the correct answer is option: 'D'.