--- title: "Projectile target distance; dependent motion" subject: "Applied Mechanics" chapter: 8 key: "semester_1/applied_mechanics:projectile-target-overshoot-undershoot-dependent-motion" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/projectile-target-overshoot-undershoot-dependent-motion site: BCT Engineering --- # Projectile target distance; dependent motion Applied Mechanics · Chapter 8: Kinematics of Particles and Rigid Body - Asked in: 2071 Shrawan, 8+2 marks - Video: [Dynamics - Lesson 7: Projectile Motion Introduction Example](https://www.youtube.com/watch?v=_q_usP2Dalk) — Jeff Hanson - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/projectile-target-overshoot-undershoot-dependent-motion - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8 ## Question 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. ## 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. ## Answer #### 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the first condition (15 deg, falls 10 m short) written? A. D + 10 = K sin 30 B. D - 10 = K sin 30 with K = v0^2/g C. D = K sin 15 D. D - 10 = K sin 15 Answer: B. Range = K sin 2α, and falling short means range = D - 10, with 2α = 30 deg. (Key point: Range formula set-up) 2. How is K found? A. Adding the equations B. Subtracting: 35 = (0.9848 - 0.5)K, so K = 72.19 m C. Dividing D by 2 D. Taking K = 35 Answer: B. Subtracting eliminates D, since (D+25) - (D-10) = 35. (Key point: Finding K) 3. Given D = 46.1 m and K = 72.19 m, how is the angle found? A. tan α = D/K B. sin 2α = D/K = 0.6385 giving α = 19.8 deg (or 70.2 deg) C. α = 15 + 40 over 2 D. cos α = D/K Answer: B. R = K sin 2α must equal D, so 2α = 39.7 deg. (Key point: Required angle) 4. Two blocks joined by a cord over a fixed pulley: x + y = L. Which is true? A. v_A = -2v_B B. v_A = -v_B and a_A = -a_B C. v_A = v_B only if masses are equal D. Cord length changes Answer: B. Differentiating the constant-length constraint gives equal and opposite rates. 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