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A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is

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

Correct answer: tan⁻¹(1/2)

Solution

At the highest point of the projectile's motion, the vertical velocity component becomes zero, and only the horizontal velocity remains. The elevation angle is given by tan⁻¹(vertical displacement/horizontal displacement). For a 45° launch angle, this simplifies to tan⁻¹(1/2).

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