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

G1 Physics · Accepted Answer

Step 1: For θ=45°, maximum height H=u²/(4g). Step 2: Horizontal distance of the highest point from the origin is R/2=u²/(2g). Step 3: If α is the elevation angle, tanα=H/(R/2)=1/2. Hence α=tan⁻¹(1/2). Final answer: B.