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:
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AT = (2u sin α) / g
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BT = (u sin 2α) / g
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CT = (u cos α) / g
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DT = (2u cos α) / g
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ET = (u^2 sin α) / g
Answer
Correct Answer: T = (2u sin α) / g
Explanation
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:
Vertical displacement at landing: 0 = u sin α * T − (1/2) g T^2.Factor T: T (u sin α − (1/2) g T) = 0 → nontrivial root T = 2u sin α / g.Thus T depends linearly on u sin α.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