Home » Mechanical Engineering » Engineering Mechanics

Projectile on an upward inclined plane — time of flight expression Given: initial speed u, angle of projection α with the horizontal, and an upward plane inclination β with the horizontal. Select the correct formula for the time of flight (until the projectile strikes the plane).

Difficulty: Medium

Correct Answer: (2u sin(α − β)) / (g cos β)

Explanation:

Setup

  • Inclined plane makes angle β with the horizontal.
  • Projectile launched at speed u and angle α with the horizontal.
  • Axes chosen: tangential to plane (‖) and normal to plane (⊥).


Components
Initial velocity along the plane: u = u cos(α − β)Initial velocity normal to the plane: u = u sin(α − β)Acceleration along the plane: a = g sin β (down the plane)Acceleration normal to the plane: a = g cos β (toward the plane)


Step-by-Step (normal motion)
Normal displacement y(t) = u sin(α − β) t − &frac{1}{2} g cosβ \, t^2 Impact with the plane occurs when y(t) returns to zero (excluding t = 0): u sin(α − β) t − &frac{1}{2} g cosβ \, t^2 = 0 \Rightarrow t = \dfrac{2u \sin(α − β)}{g \cosβ}


Common pitfalls

  • Using g instead of g cosβ for the normal component.
  • Confusing (α − β) with (α + β); the effective launch angle relative to the plane is (α − β).


Final Answer
(2u sin(α − β)) / (g cos β)

← Previous Question Next Question→

More Questions from Engineering Mechanics

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion