A ball bearing is projected vertically upwards from the ground with a velocity of 15ms-1. Calculate the time taken by the ball to return to the ground [g = ...

The ball bearing was projected upwards with an initial velocity of 15ms^-1. At the highest point, the velocity of the ball becomes zero before it starts falling back to the ground. Using the equation of motion for free-falling objects: h = ut + 1/2 gt^2 where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time. At the highest point, the height of the ball bearing is given by: h = (15)^2 / (2 × 10) = 11.25m (using v² = u² + 2gh) The time taken for the ball bearing to reach the highest point can be obtained by using the equation of motion: v = u + gt 0 = 15 - 10t t = 1.5s The total time taken for the ball bearing to reach the ground is twice the time taken to reach the highest point, which is: 2t = 2 × 1.5s = 3.0s Therefore, the time taken by the ball bearing to return to the ground is 3.0s. Hence, the answer is 3.0s.