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At some instant during its flight, a projectile has a velocity of 5 m/s directed at 30 degrees above the horizontal. How much time after this instant is the projectile moving in a direction perpendicular to this direction? (Take g = 10 m/s²)

1. 1/2 s
2. 1 s
3. sqrt(3)/2 s
4. 2 s

Correct answer: 1 s

Solution

vₓ = 5 cos30 = 5*sqrt(3)/2, v_y0 = 5 sin30 = 5/2. After time t: velocity = (vₓ, v_y0 - gt). Perpendicularity: vₓ² + v_y0*(v_y0 - gt) = 0 => 75/4 + (5/2)*(5/2 - 10t) = 0 => 25 = 25t => t = 1 s.

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