From a point 100 m above the ground, a ball is thrown horizontally with a speed of 20 m/s . Determine the velocity (both magnitude and direction) with which it strikes the ground.

Step-by-step solution

Initial horizontal velocity, u_x = 20 m/s Initial vertical velocity, u_y = 0 Vertical distance, h = 100 m Using the equation of motion for the vertical direction: v_y^2 = u_y^2 + 2gh v_y^2 = 0 + 2 9.8 100 = 1960 v_y = 1960 44.27 m/s The horizontal velocity remains constant throughout the motion, v_x = 20 m/s Magnitude of the final velocity is given by: v = v_x^2 + v_y^2 = 20^2 + 1960 = 400 + 1960 = 2360 48.58 m/s 49 m/s Let be the angle the velocity vector makes with the horizontal: = v_y v_x = 44.27 20 2.21 = ⁻¹(2