Projectile motion on level ground: time of flight For a projectile launched with initial speed u at an angle α above the horizontal (same launch and landing elevation), the time of flight T equals:

Introduction / Context:
The time of flight of a projectile on level terrain depends only on the vertical component of the initial velocity. This is a fundamental kinematics result used to compute range and maximum height.

Given Data / Assumptions:

• Initial speed u; projection angle α.
• Uniform gravitational acceleration g; no air drag.
• Landing height equals launch height.

Concept / Approach:

Vertical motion governs the ascent and descent: time to rise to peak equals time to descend back to the same elevation. Use v_y(t) = u sin α − g t and y(t) = 0 condition at landing.

Step-by-Step Solution:

Verification / Alternative check:

At α = 90°, T = 2u/g (pure vertical throw), matching the standard vertical result.

Why Other Options Are Wrong:

(b) mixes range identity; (c) and (d) use cos instead of sin; (e) is dimensionally incorrect (u^2/g is time only when accompanied by angle functions differently).

Common Pitfalls:

Applying the formula when landing elevation differs from launch; forgetting to double the ascent time.

Final Answer:

T = (2u sin α) / g