A projectile can have the same range R for two angles of projection. If t_1 and t_2 be the times of flights in the two cases, then the product of the - Physics MCQ AIEEE 2026 | ExamTest.live

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A projectile can have the same range R for two angles of projection. If t_{1} and t_{2} be the times of flights in the two cases, then the product of the two time of flights is proportional to:

Solution & Step-by-step Explanation

For a given velocity of projection u, the range R is the same for angles θ and (9 0^{\circ} - θ) .The times of flight for these angles are: t_{1} = \frac{2 u sin θ}{g} t_{2} = \frac{2 u sin ( 9 0 ^{\circ} - θ )}{g} = \frac{2 u cos θ}{g} Multiplying t_{1} and t_{2} : t_{1} t_{2} = (\frac{2 u sin θ}{g}) (\frac{2 u cos θ}{g}) t_{1} t_{2} = \frac{2}{g} (\frac{u ^{2} ( 2 sin θ cos θ )}{g}) Since R = \frac{u ^{2} s i n 2 θ}{g} : t_{1} t_{2} = \frac{2 R}{g} Thus, t_{1} t_{2} \propto R .

Try it yourself before checking the explanation above.

A projectile can have the same range R for two angles of projection. If t_{1} and t_{2} be the times of flights in the two cases, then the product of the two time of flights is proportional to:

R^{2}

1/ R^{2}

1/ R

R

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