Kinematics

JEE Main level, single correct, medium

206 of 308

Question 206 of 308

Two projectiles are fired from ground with same initial speeds from same point at angles (45∘+α)\left(45^{\circ}+\alpha\right) and (45∘−α)\left(45^{\circ}-\alpha\right) with horizontal direction. The ratio of their times of flights is

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Answer: (D) 1+tan⁡α1−tan⁡α\frac{1+\tan \alpha}{1-\tan \alpha}

θ1=45+α;θ2=45−α\theta_1=45+\alpha ; \theta_2=45-\alpha
Time of flight, T=2vsin⁡θg\mathrm{T}=\frac{2 \mathrm{v} \sin \theta}{\mathrm{g}}
T1 T2=sin⁡(45+α)sin⁡(45−α)\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\sin (45+\alpha)}{\sin (45-\alpha)}
T1 T2=12cos⁡α+12sin⁡α12cos⁡α−12sin⁡α\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}
T1 T2=cos⁡α+sin⁡αcos⁡α−sin⁡α=1+tan⁡α1−tan⁡α\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\cos \alpha+\sin \alpha}{\cos \alpha-\sin \alpha}=\frac{1+\tan \alpha}{1-\tan \alpha}

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