Two projectiles are fired with same initial speed from same point on ground at angles of (45° - α) and (45° + α), respectively, with the horizontal direction. The ratio of their maximum heights attained is :

Solution

Physics sir · 3 steps

Max height goes as sin² θ, so write each sin² with the double-angle formula and divide.

1. Height formula. For speed u at angle θ, the maximum height is H = (u² sin² θ)/2g. Both projectiles have the same u and g, so the ratio of heights is just the ratio of sin² θ.
2. Double-angle form. Use sin² θ = (1-cos 2θ)/2. For θ = 45° - α, cos 2θ = cos(90° - 2α) = sin 2α. For θ = 45° + α, cos 2θ = cos(90° + 2α) = -sin 2α.

   H₁/H₂ = sin²(45° - α)/sin²(45° + α) = ((1-sin 2α)/2)/((1+sin 2α)/2) = (1-sin 2α)/(1+sin 2α)

3. So. The lower-angle projectile rises less, and the ratio is (1-sin 2α)/(1+sin 2α), which is option (d).

The idea to keep. Angles 45° ± α give the same range but different heights. Height depends on sin² θ, and sin² θ = (1-cos 2θ)/2 turns these angles into 1 ∓ sin 2α.

Answer from the NTA final answer key. Solution by PrepWiser.