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Projectile at 30°, 61 m/s to target below

Short answer

Trajectory equation y = x tan30° – gx²/(2v₀²cos²30°) set equal to –0.75x gives x = 755.2 m, drop h = 566.4 m. Sloping distance = √(x²+h²) = 944 m. Maximum height above projection is 47.4 m; height above target = 47.4 + 566.4 = 614 m.

A particle is projected at an angle of 30° to horizontal axis with an initial velocity of 61 m/sec hit the target located at 'h' meter below the horizontal axis and having the inclined slope of ¾ downward from the axis of the target. Find the sloping distance covered by the projectile and the maximum height achieved by particle from the target.

Setting up

v₀ = 61 m/s at θ = 30°. The target lies on an incline falling 3 m for every 4 m horizontally (tan φ = 3/4, φ = 36.87°). Origin at the point of projection; the target is at (x, −0.75x).

Trajectory: y = x tan θ − g x²/(2v₀² cos²θ)

  • tan 30° = 0.5774
  • g/(2v₀² cos²30°) = 9.81/(2 × 3721 × 0.75) = 0.0017575

Target position

−0.75x = 0.5774x − 0.0017575x² 0.0017575x = 0.5774 + 0.75 = 1.3274, so x = 755.2 m h = 0.75 × 755.2 = 566.4 m below the axis.

Sloping distance

Sloping distance = √(755.2² + 566.4²) = 755.2/cos 36.87° = 944.0 m (Time of flight: t = 755.2/(61 cos 30°) = 14.30 s.)

Maximum height from the target

  • Vertical velocity: v_y = 61 sin 30° = 30.5 m/s
  • Rise above the axis: 30.5²/(2 × 9.81) = 47.41 m
  • Height above the target: 47.41 + 566.4 = 613.8 m

Answer: sloping distance ≈ 944 m; maximum height above the target ≈ 614 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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