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Projectile target distance; dependent motion

Short answer

With same v₀, range R=(v₀²/g) sin2α. Solving D-10=0.5K, D+25=0.9848K gives K=72.19 m, D=46.1 m. Required angle: sin2α=0.6385, α=19.8° or 70.2°. Dependent motion: e.g., blocks connected by cord over a pulley, constraint x+y=L, so v_A=-v_B.

A projectile is aimed at a mark on the horizontal plane through the point of projection and falls 10 m short when the angle of projection is 15° while overshoots the mark by 25 m when the inclination is 40°. Calculate the distance of the target and required angle of projection, if the velocity remains constant. Neglecting air resistance. Define dependent motion of particle with example.

Distance of the target and angle of projection

For the same speed v₀ on a horizontal plane the range is R = (v₀²/g) sin 2α. Let K = v₀²/g and D the distance of the target.

  • α = 15°: it falls 10 m short, so D − 10 = K sin 30° = 0.5K
  • α = 40°: it overshoots by 25 m, so D + 25 = K sin 80° = 0.9848K

Subtracting: 35 = (0.9848 − 0.5)K = 0.4848K, so K = 72.19 m (v₀ = √(9.81 × 72.19) = 26.6 m/s).

Distance of the target: D = 0.5(72.19) + 10 = 46.1 m (check: 0.9848 × 72.19 − 25 = 46.1 m).

Required angle: sin 2α = D/K = 46.10/72.19 = 0.6385, so 2α = 39.7° or 140.3°, giving α = 19.8° (or 70.2°).

Dependent motion of a particle

When the position of one particle depends on another because of a constraint, such as an inextensible cord, their motions are dependent. Example: two blocks A and B joined by a cord over a fixed pulley. If A moves a distance x along a table, B drops the same distance y: x + y = L, so v_A = −v_B, a_A = −a_B. For a cord passing around one movable pulley, x_A + 2y_B = L, so v_A = −2v_B.

✓ Answer checked by Claude Opus 5.5 (high effort)

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