Time of flight of a projectile (standard flat ground): If a particle is projected with initial speed u at an elevation angle α above the horizontal on level ground, what is the total time of flight until it returns to the same level (neglect air resistance)?

Introduction / Context:
The time of flight is a fundamental result of projectile motion under uniform gravity on level ground. It depends on the initial vertical component of velocity and gravity but not on the horizontal component directly.

Given Data / Assumptions:

• Initial speed = u; projection angle = α (above horizontal).
• Landing level equals launch level (flat ground).
• Acceleration due to gravity = g downward; air resistance neglected.

Concept / Approach:

Vertical motion controls flight duration. With initial vertical component u_y = u sin α, the time to rise to the top is u_y / g, and the time to descend back to the original level is the same, giving the well-known factor of 2.

Step-by-Step Solution:

Verification / Alternative check:

Symmetry of ascent and descent under constant g gives T = 2 * (u sin α / g) immediately.

Why Other Options Are Wrong:

(a) is only the ascent time; (c) and (d) incorrectly use the horizontal component; (e) is a range expression, not a time expression.

Common Pitfalls:

Confusing range and time formulas; forgetting the factor of 2; using degrees/radians inconsistently when computing sines numerically.

Final Answer:

(2 u sin α) / g