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Ball tossed upward from window; dependent motion

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.

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.

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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