A projectile can have the same range R for two angles of projection. If T_1 and T_2 be the time of flights in the two cases, then the product of the t - Physics MCQ AIEEE 2004 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 time of flights in the two cases, then the product of the two time of flights is directly proportional to:

Solution & Step-by-step Explanation

For a given velocity u, the range R is the same for complementary angles θ and (9 0^{\circ} - θ) .Time of flight for θ : T_{1} = \frac{2 u s i n θ}{g} Time of flight for (9 0^{\circ} - θ) : T_{2} = \frac{2 u s i n ( 9 0 ^{\circ} - θ )}{g} = \frac{2 u c o s θ}{g} Product of time of flights: T_{1} T_{2} = \frac{2 u sin θ}{g} \cdot \frac{2 u cos θ}{g} = \frac{2}{g} (\frac{u ^{2} ( 2 sin θ cos θ )}{g}) T_{1} T_{2} = \frac{2}{g} (\frac{u ^{2} sin 2 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 .

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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 time of flights in the two cases, then the product of the two time of flights is directly proportional to:

1/ R^{2}

1/ R

R

R^{2}

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