1. The problem statement, all variables and given/known data During baseball practice, you go up into the bleachers to retrieve a ball. You throw the ball back into the playing field at an angle of 42° above the horizontal, giving it an initial velocity of 15 m/s. If the ball is 5.3 m above the level of the playing field when you throw it, what will be the velocity of the ball when it hits the ground of the playing field? <=42° vi=15m/s h=5.3m vf=? 2. Relevant equations h=-0.5gt^2+VosinθΔt vf=vi+aΔt 3. The attempt at a solution 5.3m=-0.5(9.8)t^2+15(sin 42°)Δt t=1.02s vf=(15)(sin 42°)+(-9.8)(1.02) vf=0.041m/s vx^2 + vy^2 (square root) (15 x cos 42)^2 + 0.041^2 =11.15m/s what am i doing wrong? Answer is 18m/s.