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NEET PHYSICSMOTION IN A PLANEEasy

Question

If a body A of mass MM is thrown with a velocity vv at an angle of 3030^{\circ} to the horizontal and another body B of the same mass is thrown with the same speed at an angle of 6060^{\circ} to the horizontal. The ratio of horizontal range of A to B will be:

A

1 : 3

B

1 : 1

C

1 : √3

D

√3 : 1

Step-by-Step Solution

  1. Formula for Horizontal Range: The horizontal range RR of a projectile is given by R=v02sin2θ0gR = \frac{v_0^2 \sin 2\theta_0}{g} .
  2. Property of Complementary Angles: The horizontal range is the same for two angles of projection θ\theta and (90θ)(90^{\circ} - \theta) for the same initial speed. This is because sin2θ=sin2(90θ)=sin(1802θ)\sin 2\theta = \sin 2(90^{\circ} - \theta) = \sin(180^{\circ} - 2\theta) .
  3. Application: Given angles are 3030^{\circ} and 6060^{\circ}. Since 30+60=9030^{\circ} + 60^{\circ} = 90^{\circ}, these are complementary angles. Therefore, their ranges are equal.
  4. Mass Independence: The trajectory and range of a projectile (ignoring air resistance) are independent of the mass of the body . Thus, the ratio of ranges is 1:11:1.

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 PLANEthrownvelocityhorizontalanotherthrown

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