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JEE MAIN 2026 · Kinematics

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 31

Updated
Single correct+4 / -1Answer matches NTA’s final key

Two identical bodies, projected with the same speed at two different angles cover the same horizontal range RR. If the time of flight of these bodies are 5 s and 10 s, respectively, then the value of RR is ______ m. (Take g=10g = 10 m/s²)

Choose one

Solution

Solution

Equal ranges at the same speed mean complementary angles, θ\theta and 90∘−θ90^\circ - \theta. Then

T1T2=2usin⁡θg⋅2ucos⁡θg=2g⋅u2sin⁡2θg=2Rg.T_1T_2 = \frac{2u\sin\theta}{g}\cdot\frac{2u\cos\theta}{g} = \frac{2}{g}\cdot\frac{u^2\sin 2\theta}{g} = \frac{2R}{g}.

R=gT1T22=10×5×102=250 m.R = \frac{gT_1T_2}{2} = \frac{10 \times 5 \times 10}{2} = 250 \text{ m}.

Answer: 250 m.

Answer checked against the official answer key. The solution is our own working.

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