If t_1 and t_2 are the times of flight of two particles having the same initial velocity u and range R on the horizontal, then t_1^2+t_2^2 is equal to - Mathematics MCQ AIEEE 2004 2026 | ExamTest.live

hardMCQAIEEE 20042026Mathematics

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If t_{1} and t_{2} are the times of flight of two particles having the same initial velocity u and range R on the horizontal, then t_{1}^{2} + t_{2}^{2} is equal to

Solution & Step-by-step Explanation

For a given velocity u and horizontal range R, there are two complementary angles of projection α and β, where α + β = 9 0^{\circ} . The times of flight are: t_{1} = \frac{2 u s i n α}{g} . t_{2} = \frac{2 u s i n β}{g} = \frac{2 u s i n ( 9 0 ^{\circ} - α )}{g} = \frac{2 u c o s α}{g} . Then t_{1}^{2} + t_{2}^{2} : t_{1}^{2} + t_{2}^{2} = (\frac{2 u s i n α}{g})^{2} + (\frac{2 u c o s α}{g})^{2} . = \frac{4 u ^{2} s i n ^{2} α}{g ^{2}} + \frac{4 u ^{2} c o s ^{2} α}{g ^{2}} = \frac{4 u ^{2}}{g ^{2}} (sin^{2} α + cos^{2} α) . = \frac{4 u ^{2}}{g ^{2}} .

Try it yourself before checking the explanation above.

If t_{1} and t_{2} are the times of flight of two particles having the same initial velocity u and range R on the horizontal, then t_{1}^{2} + t_{2}^{2} is equal to

\frac{u ^{2}}{g}

\frac{4 u ^{2}}{g ^{2}}

\frac{u ^{2}}{2 g}

1

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