A man standing 31 m from the base of a vertical cliff throws a ball with a speed of 38 m/s aimed directly at a point 5 m above the base of the cliff. Question: How long does it take the ball to reach the cliff? 0.82s Neglecting air resistance and the height of the man, calculate the height above the base of the cliff at which the ball hits. 1.66m How fast is the ball moving when it reaches the cliff? 37.3m/s Here is where I am having trouble... At what time does the ball reach its largest vertical height? I am sure that I have to find the intitial vertial velocity: 31 = 38cos(x)(0.82) where x is the angle x = 5.8 degrees vy0 = 38sin(x) vy0 calculated to be 3.84 vf = v0 + at 0 = 3.84 + (-9.8)t = 0.39s time to max height i figured to be 0.39s This is the incorrect answer, what did I do wrong? Any suggestions? Thanks!