1. The problem statement, all variables and given/known data A ball is thrown at a speed of 22.2 m/s at an angle of 44.4° on to a flat level field. The ball leaves the thrower’s hand at a height of 1.88 m. (a) How long is the ball in the air, (b) how far from where the thrower stands does it hit the ground, and (c) what is the magnitude and direction of the balls velocity the instant before it strikes the ground? 2. Relevant equations 3. The attempt at a solution well first thing i did was break the projectile into it's horizontal and vertical components 22.2cos44.4 = 15.86129349 22.2sin44.4 = 15.53252616 so (a) How long is the ball in the air I used Vf² = Vi² + 2a * Δy we want Δy so Δy = (Vf² - Vi²) / 2a we know the ball starts 1.88m up so y = ((Vf² - Vi²) / 2a) + 1.88 y = ((0 - 15.53252616²)/2(-9.8)) + 1.88 this came to 2.672475824 this is the max height it reaches after this...I'm kind of stuck...I believe now i would use Δy = Vi * t + 1/2 a * t² to solve for t?