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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Here's the solution for question 2: Given: Initial speed, u = 50 m/s Angle of projection = Time to reach maximum height, t_max = 3 s We will assume the acceleration due to gravity, g = 10 m/s^2, as this value is consistent with the given information in part (i). i) Show that = (3)/(5). Step 1: Use the vertical component of velocity at maximum height. At maximum height, the vertical component of velocity (v_y) is 0. The initial vertical velocity is u_y = u . Using the kinematic equation v_y = u_y - gt_max: 0 = u - gt_max Step 2: Substitute the given values and solve for . 0 = (50) - (10)(3) 0 = 50 - 30 50 = 30 = (30)/(50) = (3)/(5) This matches the required value. ii) Find the maximum height attained by the particle. Step 1: Calculate the initial vertical velocity. u_y = u = 50 × (3)/(5) = 30 m/s Step 2: Use the kinematic equation for vertical displacement to find the maximum height (H_max). H_max = u_y t_max - (1)/(2) g t_max^2 H_max = (30)(3) - (1)/(2)(10)(3)^2 H_max = 90 - (1)/(2)(10)(9) H_max = 90 - 45 H_max = 45 m iii) The horizontal distance OA covered by the particle. Step 1: Calculate the total time of flight (T). Since the particle lands at the same horizontal level as it was projected, the total time of flight is twice the time to reach maximum height. T = 2 × t_max = 2 × 3 = 6 s Step 2: Calculate the initial horizontal velocity (u_x). First, find from = (3)/(5). Using the identity ^2 + ^2 = 1: = sqrt(1 - ^2) = sqrt(1 - ((3)/(5))^2) = sqrt(1 - (9)/(25)) = sqrt((16)/(25)) = (4)/(5) u_x = u = 50 × (4)/(5) = 40 m/s Step 3: Calculate the horizontal distance (Range R). R = u_x × T R = 40 × 6 R = 240 m iv) The direction of motion of the particle 4 seconds after projection. Step 1: Calculate the horizontal and vertical components of velocity at t = 4 s. The horizontal velocity (v_x) remains constant: v_x = u_x = 40 m/s The vertical velocity (v_y) at t = 4 s: v_y = u_y - gt v_y = 30 - (10)(4) v_y = 30 - 40 v_y = -10 m/s The negative sign indicates the particle is moving downwards. Step 2: Determine the direction of motion. The direction of motion is given by the angle that the velocity vector makes with the horizontal. = (|v_y|)/(v_x) = (|-10|)/(40) = (10)/(40) = (1)/(4) = ((1)/(4)) ≈ 14.04^ below the horizontal 3 done, 2 left today. You're making progress.