A projectile is projected with an initial velocity u, at an angle θ to the horizontal. The maximum height attained is.

Analyze the question: We need the maximum height H of a projectile launched at speed u and angle θ to the horizontal.

Concept: Only the vertical part of the motion decides the height. The vertical component of the initial velocity is u sin θ. At maximum height the vertical velocity is zero, and gravity decelerates the projectile at g.

Solution:

Use v² = u_y² − 2gH, where

• v = 0 (at the top)
• u_y = u sin θ

So:

0 = u² sin²θ − 2gH

2gH = u² sin²θ

H = u² sin²θ / (2g)

Matching the options: The options are poorly formatted, but read them as:

• A. u² sin θ / g, which has no square on sin and no factor of 2 in the denominator. Wrong.
• B. 2u² sin²θ / g, where the 2 is in the wrong place. Wrong.
• C. u sin²θ / g, which has u instead of u². Wrong.
• D. u² sin²θ / 2g. This matches our result.

Correct answer: D

Tip: Remember the three projectile formulas by their pattern:

• Maximum height: H = u² sin²θ / 2g
• Time of flight: T = 2u sinθ / g
• Range: R = u² sin2θ / g

Only the height has the 2 in the denominator, which comes from the 2g in v² = u² − 2gh. Also check dimensions: u²/g has units of length, so any option with just u (like C) cannot be a height.