This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Here are the solutions to the questions: Step 1: Calculate the time taken for stone A to reach its maximum height. Let upwards be the positive direction. Initial velocity of stone A, v_iA = +12 m/s. Final velocity of stone A at maximum height, v_fA = 0 m/s. Acceleration due to gravity, a = -9.8 m/s^2. Using the kinematic equation v_f = v_i + at: 0 = 12 + (-9.8)t_maxA 9.8 t_maxA = 12 t_maxA = (12)/(9.8) t_maxA ≈ 1.22 s The time taken for stone A to reach its maximum height is 1.22 s. Step 2: Calculate the speed, v, with which stone B is thrown downwards. When stone A reaches its maximum height, the time elapsed is t_maxA = (12)/(9.8) s. At this instant, the speed of stone B is 3v. For stone B, let downwards be the positive direction for its initial velocity, or continue with upwards as positive and use negative signs. Let's stick to upwards as positive. Initial velocity of stone B, v_iB = -v. Final velocity of stone B at time t_maxA, v_fB = -3v. Acceleration due to gravity, a = -9.8 m/s^2. Using the kinematic equation v_f = v_i + at: -3v = -v + (-9.8)((12)/(9.8)) -3v = -v - 12 -2v = -12 v = 6 m/s The speed v with which stone B is thrown downwards is 6 m/s. Step 3: Calculate the height h. Stone A passes its initial position on its way down at a time t_returnA = 2 × t_maxA. t_returnA = 2 × (12)/(9.8) = (24)/(9.8) s At this instant, stone B hits the ground. For stone B: Initial velocity, v_iB = -v = -6 m/s (since v is a speed, and it's thrown downwards). Time, t = t_returnA = (24)/(9.8) s. Acceleration, a = -9.8 m/s^2. Displacement, y_B = -h. Using the kinematic equation y = v_i t + (1)/(2)at^2: -h = (-6)((24)/(9.8)) + (1)/(2)(-9.8)((24)/(9.8))^2 -h = -(144)/(9.8) - 4.9 ((576)/(9.8^2)) -h = -(144)/(9.8) - (576)/(2 × 9.8) -h = -(144)/(9.8) - (288)/(9.8) -h = -(432)/(9.8) h = (432)/(9.8) h ≈ 44.08 m The height h is 44.08 m. Step 4: Sketch velocity-time graphs for the complete motions of stones A and B. Both graphs will be straight lines with a constant slope of -9.8 m/s^2 (acceleration due to gravity). For Stone A: Initial velocity: v_iA = 12 m/s. Time to reach maximum height: t_maxA ≈ 1.22 s. At this time, velocity is 0 m/s. Time to return to initial height: t_returnA ≈ 2.45 s. At this time, velocity is -12 m/s. Total time to hit the ground (displacement y_A = -h = -44.08 m): v_fA^2 = v_iA^2 + 2a y_A = (12)^2 + 2(-9.8)(-44.08) ≈ 1007.97 v_fA ≈ -31.75 m/s t_totalA = v_fA - v_iAa = (-31.75 - 12)/(-9.8) ≈ 4.46 s The graph for stone A starts at (0, 12), passes through (1.22, 0), (2.45, -12), and ends at approximately (4.46, -31.75). For Stone B: Initial velocity: v_iB = -v = -6 m/s. Time to hit the ground (when A returns to initial position): t_returnA ≈ 2.45 s. Final velocity when hitting the ground: v_fB = v_iB + at_returnA = -6 + (-9.8)((24)/(9.8)) = -6 - 24 = -30 m/s The graph for stone B starts at (0, -6) and ends at (2.45, -30). Sketch of Velocity-Time Graphs: A velocity-time graph would show: X-axis: Time (s) Y-axis: Velocity (m/s) Graph for Stone A (Line A): A straight line starting at (0, 12), sloping downwards, crossing the time axis at t = 1.22 s* (time to reach maximum height), continuing to t ≈ 2.45 s where its velocity is -12 m/s, and ending at approximately (4.46, -31.75). Graph for Stone B (Line B): A straight line starting at (0, -6 m/s*), sloping downwards parallel to Line A, and ending at approximately (2.45, -30). Labels: Mark 1.22 s on the time axis, indicating t_maxA. Mark -6 m/s on the velocity axis, indicating the initial velocity of stone B. Label the lines clearly as "Stone A" and "Stone B". 3 done, 2 left today. You're making progress.