The time of flight for a projectile motion is given by the expression

Analyze the question: We need the total time a projectile stays in the air when launched with speed U at angle θ above the horizontal (landing at the same level it was launched from).

Concept: Only the vertical motion decides how long the projectile stays in the air. Gravity acts vertically, so we use the equation of motion for the vertical component.

Solution:

1. The vertical component of the initial velocity is U sin θ.
2. The vertical displacement is s = 0 when the projectile lands at the same level it started from.
3. Use s = ut + ½at², with u = U sin θ and a = −g:

0 = (U sin θ)t − ½gt²

4. Factorise:

t(U sin θ − ½gt) = 0

5. t = 0 is the moment of launch. The other solution gives the landing time:

U sin θ = ½gt, so t = 2U sin θ / g

Check by another route: The time to reach maximum height is U sin θ / g (where vertical velocity = 0). The descent takes the same time, so the total is 2 × (U sin θ / g) = 2U sin θ / g. ✔

Why the other options are wrong:

• A is only the time to reach maximum height.
• B is half of that, which has no physical meaning here.
• D uses sin 2θ, which belongs to the range formula (R = U² sin 2θ / g), not the time of flight.

**Correct answer: C.

Exam tip: Remember "T = 2 × time to the top". Time of flight is 2U sin θ / g, maximum height is U² sin²θ / 2g, and range is U² sin 2θ / g. Only the range formula contains sin 2θ.