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For a projectile projected at angles (45∘−θ)(45^{\circ}-\theta) and (45∘+θ)(45^{\circ}+\theta), the horizontal ranges described by the projectile are in the ratio of:

A

1:1

B

2:3

C

1:2

D

2:1

Step-by-Step Solution

  1. Formula for Horizontal Range: The horizontal range RR of a projectile projected with initial velocity v0v_0 at an angle θ0\theta_0 is given by R=v02sin⁡2θ0gR = \frac{v_0^2 \sin 2\theta_0}{g} .
  2. Analyze First Angle: Let θ1=45∘−θ\theta_1 = 45^{\circ} - \theta. R1=v02sin⁡2(45∘−θ)g=v02sin⁡(90∘−2θ)gR_1 = \frac{v_0^2 \sin 2(45^{\circ} - \theta)}{g} = \frac{v_0^2 \sin (90^{\circ} - 2\theta)}{g} Using the identity sin⁡(90∘−x)=cos⁡x\sin(90^{\circ} - x) = \cos x, we get R1=v02cos⁡2θgR_1 = \frac{v_0^2 \cos 2\theta}{g}.
  3. Analyze Second Angle: Let θ2=45∘+θ\theta_2 = 45^{\circ} + \theta. R2=v02sin⁡2(45∘+θ)g=v02sin⁡(90∘+2θ)gR_2 = \frac{v_0^2 \sin 2(45^{\circ} + \theta)}{g} = \frac{v_0^2 \sin (90^{\circ} + 2\theta)}{g} Using the identity sin⁡(90∘+x)=cos⁡x\sin(90^{\circ} + x) = \cos x, we get R2=v02cos⁡2θgR_2 = \frac{v_0^2 \cos 2\theta}{g}.
  4. Conclusion: Since R1=R2R_1 = R_2, the ranges are equal. The ratio is 1:1. This confirms the principle that for elevations which exceed or fall short of 45∘45^{\circ} by equal amounts, the ranges are equal .
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