The ratio of maximum range to maximum height of a projectile is 4. Determine the angle of projection.

Analyze the question

We are told the ratio of range R to maximum height H is 4, and we need the angle of projection θ.

Concept

For a projectile launched with speed u at angle θ:

• Range: R = u² sin2θ / g = 2u² sinθ cosθ / g
• Maximum height: H = u² sin²θ / (2g)

(The "maximum range" here means the horizontal range of the projectile. The range is greatest at 45°, but the question uses it as the ratio R/H.)

Solution

R/H = [2u² sinθ cosθ / g] ÷ [u² sin²θ / 2g]

R/H = (2 sinθ cosθ × 2) / sin²θ = 4 cosθ / sinθ = 4 cotθ

Set R/H = 4:

4 cotθ = 4 cotθ = 1 tanθ = 1 θ = 45°

Check: At 45°, sinθ = cosθ, so R = u²/g and H = u²/(4g). R/H = 4 ✓.

Answer: Option C (45°)

Tip: Remember the general result R/H = 4/tanθ. If a question gives R/H = 4, the angle is automatically 45°, because tan45° = 1. Memorise R = u² sin2θ/g and H = u² sin²θ/2g, as they appear often in JAMB and WASSCE.