I found the initial velocity in the x direction to be 14cos(225)=5.14 m/s and the initial velocity in the y direction to be 14sin(225)=-13.02 m/s. The equation I am trying to plug it into is V1y(SQ)=V0y(SQ)+2aΔt
A woman throws a ball at a vertical wall d = 3.2 m away. The ball is h = 1.6 m above ground when it leaves the woman's hand with an initial velocity of 14 m/s at 45°. When the ball hits the wall, the horizontal component of its velocity is reversed; the vertical component remains unchanged...