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A projectile is fired at an angle of 45∘45^{\circ} with the horizontal. The elevation angle α\alpha of the projectile at its highest point as seen from the point of projection is:

A

60∘60^{\circ}

B

tan⁡−1(12)\tan^{-1}\left(\frac{1}{2}\right)

C

tan⁡−1(32)\tan^{-1}\left(\frac{\sqrt{3}}{2}\right)

D

45∘45^{\circ}

Step-by-Step Solution

  1. Geometry of the Path: Let the projectile be fired with initial velocity v0v_0 at an angle θ=45∘\theta = 45^\circ. The highest point (PP) of the trajectory has coordinates (x,y)=(R/2,H)(x, y) = (R/2, H), where RR is the horizontal range and HH is the maximum height.
  2. Angle of Elevation: The angle of elevation α\alpha is the angle made by the position vector of point PP with the horizontal. From trigonometry: tan⁡α=yx=HR/2=2HR\tan \alpha = \frac{y}{x} = \frac{H}{R/2} = \frac{2H}{R}
  3. Formulas:
  • Maximum Height: H=v02sin⁡2θ2gH = \frac{v_0^2 \sin^2 \theta}{2g} .
  • Horizontal Range: R=v02sin⁡2θg=2v02sin⁡θcos⁡θgR = \frac{v_0^2 \sin 2\theta}{g} = \frac{2v_0^2 \sin \theta \cos \theta}{g} .
  1. Substitution: tan⁡α=2(v02sin⁡2θ2g)2v02sin⁡θcos⁡θg=sin⁡2θ2sin⁡θcos⁡θ=12tan⁡θ\tan \alpha = \frac{2 \left( \frac{v_0^2 \sin^2 \theta}{2g} \right)}{\frac{2v_0^2 \sin \theta \cos \theta}{g}} = \frac{\sin^2 \theta}{2 \sin \theta \cos \theta} = \frac{1}{2} \tan \theta
  2. Calculation: Given θ=45∘\theta = 45^\circ: tan⁡α=12tan⁡45∘=12(1)=12\tan \alpha = \frac{1}{2} \tan 45^\circ = \frac{1}{2}(1) = \frac{1}{2} α=tan⁡−1(12)\alpha = \tan^{-1}\left(\frac{1}{2}\right)
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