A student throws a ball with a speed of 15.0 m.s-1 at an angle of 40.0° above the horizontal directly toward a wall as shown below. The wall is 10.0 m from the release point of the ball. Now I did figure this out. The vertical component of velocity (initial) is v= v(inital) sin angle, and in the vertical direction the ball has the constant acceleration due to gravity. So v = v(initial) sin angle - gt should be true for any given time. I can't figure out how far above the release point does the ball hit the wall?