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The speed of a projectile at its maximum height is sqrt(2/5) times its speed at half of its maximum height. Find the angle of projection.

1. 30 deg
2. 45 deg
3. tan⁻¹(2/3)
4. 60 deg

Correct answer: 60 deg

Solution

Maximum height H = u²*sin²(theta)/(2g). At half max height h = H/2: vy² = u²*sin²(theta) - 2g*(H/2) = u²*sin²(theta)/2. Speed at half height: v_half² = u²*cos²(theta) + u²*sin²(theta)/2. Speed at top: v_top = u*cos(theta). Given v_top = sqrt(2/5)*v_half: u²*cos²(theta) = (2/5)*(u²*cos²(theta) + u²*sin²(theta)/2). Let c = cos²(theta): c = (2/5)*(c + (1-c)/2) = (1/5)*(c+1). 5c = c+1 => 4c = 1 => c = 1/4 => cos(theta) = 1/2 => theta = 60 deg.

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