A body is projected with an initial velocity U at an angle O to the horizontal. The time taken by it to reach its maximum height is given by the expression

Analyze the question: We need the time for a projectile to reach its maximum height (not the total time of flight).

Concept: Only the vertical component of motion decides how high the body rises. At maximum height, the vertical velocity is momentarily zero.

Solution:

1. Vertical component of the initial velocity: u_y = u sinθ
2. Gravity acts downward, so the acceleration is a = −g
3. At maximum height, the final vertical velocity v = 0
4. Use v = u_y + at:

0 = u sinθ − gt

gt = u sinθ

t = u sinθ / g

Checking the other options:

• A, 2u sinθ / g, is the total time of flight (up and down), which is twice the time to reach the top.
• B, u² sin²θ / 2g, is the maximum height H. It is a distance, not a time.
• D, u sinθ / 2g, has no physical meaning here.

Exam tip: Remember the trio for projectile motion:

• Time to maximum height: T = u sinθ / g
• Total time of flight: 2u sinθ / g (double the time to the top)
• Maximum height: u² sin²θ / 2g

Check the units: time must have units of seconds. (m/s) ÷ (m/s²) = s, so C is dimensionally correct.