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Two projectiles are fired from ground with same initial speeds from same point at angles \(\left(45^\circ +\alpha \right…

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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

[JEE Main 2025, 7 Apr (Shift 1)]

a

\(1\)

b

\(\frac{1-\tan \alpha }{1+\tan \alpha }\)

c

\(\frac{1+\sin 2\alpha }{1-\sin 2\alpha }\)

d

\(\frac{1+\tan \alpha }{1-\tan \alpha }\)

✓ Correct answer: d)

\(\frac{1+\tan \alpha }{1-\tan \alpha }\)

Explanation

Angles of projection
\({\theta }_{1}=45+\alpha\)
\({\theta }_{2}=45-\alpha\)
Time of flight formula
\(T=\frac{2v\sin \theta }{g}\)
Ratio of times of flight
\(\frac{{T}_{1}}{{T}_{2}}=\frac{\sin (45+\alpha )}{\sin (45-\alpha )}\)
Using trigonometric identities
\(\frac{{T}_{1}}{{T}_{2}}=\frac{\frac{1}{\sqrt{2}}\cos \alpha +\frac{1}{\sqrt{2}}\sin \alpha }{\frac{1}{\sqrt{2}}\cos \alpha -\frac{1}{\sqrt{2}}\sin \alpha }\)
Simplifying
\(\frac{{T}_{1}}{{T}_{2}}=\frac{\cos \alpha +\sin \alpha }{\cos \alpha -\sin \alpha }\)
Final ratio
\(\frac{{T}_{1}}{{T}_{2}}=\frac{1+\tan \alpha }{1-\tan \alpha }\)

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