A projectile fired with initial velocity, u will attain its maximum height H when the angle of projection θ is

Analyze the question

For a fixed initial speed u, we want the angle θ that gives the greatest height H. (Note that "450°" in option B is likely a misprint of 45°.)

Concept

The vertical component of the initial velocity is u sinθ. The maximum height depends only on this component:

H = u² sin²θ / (2g)

Solution

• H is largest when sin²θ is largest.
• The largest value of sinθ is 1, which happens at θ = 90°.
• At 90°, H = u²/(2g), the greatest height possible for that speed.

Checking the options:

• 30°: sin²θ = 0.25
• 60°: sin²θ = 0.75
• 75°: sin²θ ≈ 0.933
• 90°: sin²θ = 1 (maximum)
• 450° (if read literally): sin 450° = sin 90° = 1, so this would give the same height mathematically. But it is not a sensible angle of projection, and it is almost certainly a typo for 45°, where sin²θ = 0.5.

The correct answer is therefore 90°, option E.

Tip

Don't confuse the two cases:

• Maximum height: θ = 90° (fired straight up).
• Maximum range: θ = 45° (on level ground).

Read the question carefully to see whether it asks about height or range.