Projectile motion on a horizontal plane – time of flight For a projectile launched with initial speed u at an angle θ above the horizontal (neglect air resistance), the total time of flight t back to the launch level is:

Introduction / Context:The time of flight of a projectile returning to its original elevation depends on the vertical component of its initial velocity and gravity. This is a staple result in kinematics.

Given Data / Assumptions:

• Initial speed u, launch angle θ above horizontal.
• Constant gravitational acceleration g downward.
• Projectile lands at the same vertical level as launch point.

Concept / Approach:Analyze vertical motion. The vertical component of the initial velocity is u sin θ. The projectile rises until vertical velocity becomes zero, then descends symmetrically, making the total time twice the time to reach the peak.

Step-by-Step Solution:

Verification / Alternative check:Use displacement equation y = u sin θ * t − (1/2) g t^2 and set y = 0 (besides t = 0). Solving yields t = (2 u sin θ)/g, confirming the result.

Why Other Options Are Wrong:

• (u sin θ)/g: only the ascent time.
• (u^2 sin 2θ)/g: related to horizontal range, not time.
• (2 u cos θ)/g: uses the horizontal component incorrectly.

Common Pitfalls:Forgetting the factor of 2 or using sin 2θ (a range formula) in the time expression.

Final Answer:t = (2 u sin θ) / g