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 for the projectile motion problem. Given: Height of the cliff H = 50 \, m Initial horizontal velocity u_x = 10 \, m/s Acceleration due to gravity g = 10 \, m/s^2 Initial vertical velocity u_y = 0 \, m/s (since the ball is projected horizontally) a) The time taken to reach the ground (t) Step 1: Use the kinematic equation for vertical motion. The vertical displacement is H. Since the initial vertical velocity is zero, the equation is: H = u_y t + (1)/(2)gt^2 Step 2: Substitute the given values. 50 \, m = (0 \, m/s)t + (1)/(2)(10 \, m/s^2)t^2 50 = 5t^2 Step 3: Solve for t. t^2 = (50)/(5) t^2 = 10 t = sqrt(10) \, s t ≈ 3.162 \, s The time taken to reach the ground is 3.16 s. b) The vertical component of the velocity when the ball hits the ground (v_y) Step 1: Use the kinematic equation for final vertical velocity. v_y = u_y + gt Step 2: Substitute the values for u_y, g, and t. v_y = 0 \, m/s + (10 \, m/s^2)(sqrt(10) \, s) v_y = 10sqrt(10) \, m/s v_y ≈ 31.62 \, m/s The vertical component of the velocity when the ball hits the ground is 31.6 m/s. c) The distance from the foot of the cliff to where the ball hits the ground (horizontal range, R) Step 1: Use the formula for horizontal distance. Since there is no horizontal acceleration, the horizontal velocity remains constant. R = u_x t Step 2: Substitute the values for u_x and t. R = (10 \, m/s)(sqrt(10) \, s) R = 10sqrt(10) \, m R ≈ 31.62 \, m The distance from the foot of the cliff to where the ball hits the ground is 31.6 m.