A particle is projected vertically upwards. The ratio of its velocity at t = 2 s to its velocity at t = 4 s is 3 2 . The ratio of the distance covered by the particle in the 3^ rd second to the distance covered in the 6^ th second is x 5 .

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A particle is projected vertically upwards. The ratio of its velocity at t = 2 s to its velocity at t = 4 s is 3 2 . The ratio of the distance covered by the particle in the 3^ rd second to the distance covered in the 6^ th second is x 5 . The value of x is (Take g = 10 m s ⁻² )

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Correct answer: D. 11

Let the initial velocity of the particle be u . The velocity of the particle at any time t is given by v = u - gt . At t = 2 s , v₂ = u - 20 At t = 4 s , v₄ = u - 40 According to the given condition: v₂ v₄ = 3 2 u - 20 u - 40 = 3 2 2(u - 20) = 3(u - 40) 2u - 40 = 3u - 120 u = 80 m s ⁻¹ The distance covered by a particle in the n^ th second is given by: S_n = u - g 2 (2n - 1) For the 3^ rd second ( n = 3 ): S₃ = 80 - 10 2 (2 3 - 1) S₃ = 80 - 5(5) = 80 - 25 = 55 m For the 6^ th second ( n = 6 ): S₆ = 80 - 10 2 (2 6 -

A particle is projected vertically upwards. The ratio of its velocity at t = 2 s to its velocity at t = 4 s is 3 2 . The ratio of the distance covered by the particle in the 3^ rd second to the distance covered in the 6^ th second is x 5 . The value of x is (Take g = 10 m s ⁻² )

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