A projectile attains its maximum height when the angle of projection is?

Analyze the question: The question asks at which angle of projection a projectile reaches its maximum height. It does not ask about maximum range, which is a different thing and the common trap here.

Concept: For a projectile launched with speed u at angle θ to the horizontal:

• Vertical component of velocity: u sin θ
• At the top, the vertical velocity is 0, so using v² = u² − 2gH:

0 = u² sin²θ − 2gH

H = u² sin²θ / 2g

Solution:

For a fixed launch speed u, H depends only on sin²θ. It is largest when sin θ is largest, and sin θ has a maximum value of 1, at θ = 90°.

Check the options:

• 30°: sin²θ = 0.25
• 45°: sin²θ = 0.50
• 60°: sin²θ = 0.75
• 75°: sin²θ ≈ 0.93
• 90°: sin²θ = 1 (the greatest)

So the maximum height is H = u²/2g, which occurs when the projectile is thrown vertically upward at 90°.

Common trap: 45° gives the maximum range (R = u² sin 2θ / g, largest when 2θ = 90°). It does not give the maximum height. Many students pick B because they confuse the two.

Exam tip:

• Maximum height → sin²θ is largest → 90°
• Maximum range → sin 2θ is largest → 45°

Always check whether the question says "height" or "range".