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.
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You're on a roll — Here are the solutions to your questions: 13. a) State the nature of the motion of the particle between: (i) A and B The displacement remains constant. Nature of motion: Stationary or at rest. (ii) B and C The displacement is increasing, and the slope of the graph is increasing. Nature of motion: Accelerating or moving with increasing velocity. (iii) C and D The displacement is decreasing linearly, indicating a constant negative slope. Nature of motion: Moving with constant negative velocity or moving back towards the origin with constant speed. 13. b) A ball is thrown horizontally from top of a vertical tower and strikes the ground at a point 50m from the bottom of the tower. Given that the height of the tower is 45m. Determine the: Given: Height of tower, h = 45 m Horizontal distance, x = 50 m Acceleration due to gravity, g = 9.8 m/s^2 (i) Time taken by the ball to hit the ground. Step 1: Consider the vertical motion. Initial vertical velocity, u_y = 0 m/s (since it's thrown horizontally). Vertical displacement, s_y = h = 45 m. Use the kinematic equation: s_y = u_y t + (1)/(2)gt^2. 45 m = (0 m/s)t + (1)/(2)(9.8 m/s^2)t^2 45 = 4.9t^2 Step 2: Solve for t. t^2 = (45)/(4.9) t^2 ≈ 9.18367 t = sqrt(9.18367) t ≈ 3.03 s The time taken by the ball to hit the ground is 3.03 s. (ii) Initial horizontal velocity of the ball. Step 1: Consider the horizontal motion. Horizontal displacement, s_x = 50 m. Time of flight, t ≈ 3.03 s (from part i). Horizontal velocity, u_x, is constant. Use the equation: s_x = u_x t. 50 m = u_x (3.03 s) Step 2: Solve for u_x. u_x = 50 m3.03 s u_x ≈ 16.5 m/s The initial horizontal velocity of the ball is 16.5 m/s. (iii) Vertical velocity of the ball just before striking the ground. Step 1: Consider the vertical motion. Initial vertical velocity, u_y = 0 m/s. Time of flight, t ≈ 3.03 s. Acceleration, g = 9.8 m/s^2. Use the kinematic equation: v_y = u_y + gt. v_y = 0 m/s + (9.8 m/s^2)(3.03 s) Step 2: Solve for v_y. v_y ≈ 29.694 m/s v_y ≈ 29.7 m/s The vertical velocity of the ball just before striking the ground is 29.7 m/s. Send me the next one 📸