A body of mass 10 kg is projected at an angle of 45° with the horizontal. The trajectory of the body is observed to pass through a point (20, 10). If T is the time of flight, then its momentum vector, at time t = T/√2 is [Take g = 10 m/s²]
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Question

A body of mass 10 kg is projected at an angle of 45° with the horizontal. The trajectory of the body is observed to pass through a point (20, 10). If T is the time of flight, then its momentum vector, at time t = T/√2 is [Take g = 10 m/s²]

Options

G1 Physics · Accepted Answer

For θ = 45°, the trajectory is y = x - gx²/u². Since it passes through (20,10): 10 = 20 - 10(20²)/u², giving u² = 400 and u = 20 m/s. The time of flight is T = 2u sin45°/g = 2√2 s, so t = T/√2 = 2 s. Then vₓ = 10√2 and vᵧ = 10√2 - 20. For m = 10 kg, p⃗ = 100√2î + (100√2 - 200)ĵ kg·m/s. Hence option D.