engineering physics MCQ #3667

A body is projected at 45° to the horizontal. What is its range?

engineering physics MCQ #3667

  1. Question 1

    Q1. A body is projected at 45° to the horizontal. What is its range?

    • A) v² / g
    • B) v²sin(2θ) / g
    • C) v²sin(θ) / g
    • D) v²cos(θ) / g

    Answer: v²sin(2θ) / g

    Explanation: Range = v²sin(2θ) / g. At 45°, sin(2θ) = 1, so range = v² / g, using the range formula.