--- title: "Ball tossed upward from window; dependent motion" subject: "Applied Mechanics" chapter: 8 key: "semester_1/applied_mechanics:ball-tossed-vertical-window-dependent-motion" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/ball-tossed-vertical-window-dependent-motion site: BCT Engineering --- # Ball tossed upward from window; dependent motion Applied Mechanics · Chapter 8: Kinematics of Particles and Rigid Body - Asked in: 2072 Chaitra, 8+2 marks - Video: [A ball is thrown vertically upward from the 12  m level with an initial velocity of 18 m / s. At ...](https://www.youtube.com/watch?v=Yppe7F7OD20) — PW Solutions - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/ball-tossed-vertical-window-dependent-motion - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8 ## Question A ball is tossed with velocity of 10 m/s directed vertically upward from a window located 20 m above the ground. Knowing that the acceleration of the ball is constant and equal to 9.81 m/s² downward, determine: i) The velocity v and the elevation y of the ball above the ground at any time t. ii) The highest elevation reached by the ball and the corresponding value of t. iii) The time when the ball will hit the ground and the corresponding velocity. What do you mean by dependent motion? Explain with example. ## Short answer i) v=10-9.81t m/s, y=20+10t-4.905t² m. ii) Highest: at t=1.02 s, y=25.1 m. iii) Hits ground at t=3.28 s, v=-22.2 m/s (down). Dependent motion: e.g., two blocks linked by a cord over a pulley, with constraint x_A + y_B = L, so v_A = -v_B. ## Answer #### Ball tossed upward from the window Take the ground as origin, upward positive: y₀ = 20 m, v₀ = +10 m/s, a = −9.81 m/s². **(i) Velocity and elevation at any time t** - v = v₀ + at = **10 − 9.81t** (m/s) - y = y₀ + v₀t + ½at² = **20 + 10t − 4.905t²** (m) **(ii) Highest point:** v = 0 so 10 − 9.81t = 0, giving **t = 1.019 s**. y_max = 20 + 10(1.019) − 4.905(1.019)² = 20 + 10.19 − 5.09 = **25.10 m** above the ground. **(iii) Hitting the ground:** y = 0: 4.905t² − 10t − 20 = 0, so t = [10 + √(100 + 392.4)]/9.81 = (10 + 22.19)/9.81 = **3.281 s** Velocity: v = 10 − 9.81(3.281) = **−22.19 m/s**, i.e. 22.19 m/s downward (check: v² = 10² + 2(9.81)(20) = 492.4, v = 22.19 m/s). #### Dependent motion Two particles have dependent motion when their positions are linked by a constraint such as an inextensible cord. Example: a cord of length L passes over a fixed pulley and holds block A on a table at one end and a hanging block B at the other: x_A + y_B = L. Then v_A = −v_B and a_A = −a_B. With a movable pulley, x_A + 2y_B = L and v_A = −2v_B. Only one coordinate is independent. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With the ground as origin and up positive, which gives y(t)? A. y = 10t - 4.905t^2 B. y = 20 + 10t - 9.81t^2 C. y = 20 + 10t - 4.905t^2 D. y = 20 - 10t + 4.905t^2 Answer: C. y = y0 + v0 t + 1/2 a t^2 with y0 = 20, v0 = +10, a = -9.81. (Key point: Elevation equation) 2. How is the highest elevation found? A. Set a = 0 B. Set y = 0 C. Set v = 0, giving t = 1.019 s and y = 25.10 m D. Set t = 1 s Answer: C. At the top the ball momentarily has zero velocity. (Key point: Highest point condition) 3. Which is correct for the time of hitting the ground? A. Solve 4.905t^2 - 10t - 20 = 0 and take the positive root, 3.281 s B. t = 2v0/g = 2.04 s C. Take the negative root D. t = 20/10 = 2 s Answer: A. Set y = 0 and keep the physical positive root; the velocity is then -22.19 m/s. (Key point: Hitting the ground) 4. For cord over a fixed pulley holding A on a table and B hanging, x_A + y_B = L. What follows? A. v_A = -v_B B. v_A = -2v_B C. v_A = v_B D. v_A = 0 Answer: A. Differentiating gives v_A + v_B = 0: equal speeds, opposite coordinate directions. 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