1. The problem statement, all variables and given/known data First part of the question (SOLVED) A football player kicks a football so that the angle of incidence is 50 DEG and the initial magnitude of velocity of the ball is 15 m/s. Find the: a) Balls maximum height = 6.7 m b) Time of flight = 2.3 s c) time when the ball reaches the maximum height = 1.2 s d) horizontal distance = 23 m e) velocity at impact = 15 m/s [50 DEG below horizontal] Second part of the question (Need help) Now imagine that the situation is exactly the same, except this time, the ball is caught by a player when it is at a height of 1.5 m above the ground. Determine the horizontal range of the ball, the length of time that it was in the air, and the final velocity. 2. The attempt at a solution Let [up] be the positive y-direction and [forward] be the positive x-direction. Vertically V1y = +11.49 m.s Ay = -9.8 m/s Total y displacement = 1.5m Using the vertical component to find time: Displacement y = V1y (change in time) + 1/2 Ay (change in time)^2 1.5 m = (+11.49 m/s)(change in time) + 1/2 (-9.8 m/s^2)(change in time)^2 There is a quadratic equation to solve in this equation so it has been rearranged (9.8 m/s^2)(change in time)^2 + (-11.49 m/s)(change in time) + 1.5m = 0 Now my workbook gives the rearrangement differently but doesn't explain why and this is where I need help. My workbook says the rearrangement should look like this: (4.9 m/s^2)(change in time)^2 + (-11.49 m/s)(change in time) + 1.5m = 0 Can someone please explain why the acceleration was divided in two?