A particle is released from a certain height H = 400 m. Due to the wind the particle gathers the horizontal velocity component vₓ = ay where a = √5 s⁻¹ and y is the vertical displacement of the particle from point of release, then the horizontal drift of the particle when it strikes the ground is
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Question

A particle is released from a certain height H = 400 m. Due to the wind the particle gathers the horizontal velocity component vₓ = ay where a = √5 s⁻¹ and y is the vertical displacement of the particle from point of release, then the horizontal drift of the particle when it strikes the ground is

Options

G1 Physics · Accepted Answer

Let y be downward displacement from the release point. Then y = ½gt² and vₓ = ay = (ag/2)t². The horizontal drift is x = ∫₀ᵀvₓdt = agT³/6. For H = 400 m and g = 10 m/s², T = √(2H/g) = 4√5 s. With a = √5 s⁻¹, x = (√5)(10)(4√5)³/6 = 8000/3 m ≈ 2.67 km. Hence option A.