Asked in UHS MDCAT 2010 2010Moderate

Time of projectile's flight is:

Correct answer: B. 2v0 sin(θ) / g

  • A. v0 sin(θ) / g
  • B. 2v0 sin(θ) / g
  • C. v0 cos(θ) / g
  • D. 2v0 cos(θ) / g

Explanation

The correct formula for the time of flight of a projectile is 2v0 sin(θ) / g, where v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. This formula considers the total time for the projectile to ascend and descend back to the same vertical level. Option A calculates maximum height, Option C calculates the horizontal component of velocity, and Option D inaccurately applies the horizontal component in a time-related formula.

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About Projectile Motion

Projectile motion combines uniform horizontal motion with vertically accelerated motion under gravity, assuming air resistance is neglected. Questions use the components of initial velocity to find time of flight, maximum height, horizontal range and position, while distinguishing projectile motion from general circular or one dimensional motion.

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