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NEET PHYSICSMOTION IN A PLANEEasy

Question

For a projectile projected at angles (45θ)(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=v02sin2θ0gR = \frac{v_0^2 \sin 2\theta_0}{g} .
  2. Analyze First Angle: Let θ1=45θ\theta_1 = 45^{\circ} - \theta. R1=v02sin2(45θ)g=v02sin(902θ)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(90x)=cosx\sin(90^{\circ} - x) = \cos x, we get R1=v02cos2θgR_1 = \frac{v_0^2 \cos 2\theta}{g}.
  3. Analyze Second Angle: Let θ2=45+θ\theta_2 = 45^{\circ} + \theta. R2=v02sin2(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)=cosx\sin(90^{\circ} + x) = \cos x, we get R2=v02cos2θgR_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 4545^{\circ} by equal amounts, the ranges are equal .

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A PLANE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOTION IN A PLANEprojectileprojectedanglescircthetacirctheta

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