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RRB Junior Engineer / DMS / CMA 2019 · Question 3544 of 3719

'If 'θ' is angle of projection and 'u' is velocity of projection for a projectile, then its horizontal range is given by-

  1. AR = (u 2sin2θ) / gCorrect
  2. BR = (u 2cosθ) / g
  3. CR = (u 2cos2θ) / g
  4. DR = (u 2sinθ) / g

Answer: A. R = (u 2sin2θ) / g

Explanation

Concept: Projectile motion: Projectile motion is the motion of an object projected into the air , under only the acceleration of gravity. The object is called a projectile , and its path is called its trajectory . Initial Velocity: The initial velocity can be given as x components and y components. Component of initial velocity in x-direction, (u x) = ucosθ Component of initial velocity in the y-direction, (u y) = usinθ In the case of projectile motion , we can see a free-fall motion of a body on a parabolic path with constant velocity. If a body is thrown at a certain angle then during its movement, we get two components of velocity as given below. And thus, the range of a projectile is the displacement of a particle along the x- axis and can be given as: The range of the projectile, Whereas the time of flight is the total time for which projectile stayed in the air. Time of flight for the projectile, The angle of projection = θ Initial velocity = u Gravitational acceleration = g Time of flight = t Range of projectile = R Explanation: As given above, v = initial velocity of projectile θ = angle with x-axis Time of flight for the projectile , And the range of the projectile, R = ( ) ucosθ × t t = ( ) ​ g 2vsinθ R = ( ) vcosθ × t Thus, by comparing the above two equation range of projectile can also be modified as But (By using trigonometric relation)

Source: RRB JE 2019 (CBT 2) (ME) Previous Year Paper (31 Aug 2019) — prepp.in solved-paper PDF (answer key with explanations) · reliable-secondary

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