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A projectile is launched at an angle of 45° to the horizontal. The angle of elevation of the projectile from the point of launch when it is at its maximum height is

1. 60°
2. tan⁻¹(1/2)
3. tan⁻¹(√3/2)
4. 45°

Correct answer: tan⁻¹(1/2)

Solution

For launch angle 45 deg, range R = u^2/g and max height H = u^2/(4g) at horizontal distance R/2 = u^2/(2g). The elevation angle from launch point satisfies tan(theta) = H/(R/2) = (u^2/4g)/(u^2/2g) = 1/2, so theta = tan^-1(1/2).

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