The wall is ¼ of the range away.The maximum height is ½ of the range away.So it takes as long to get from the throw point to the wall as it takes to get from the wall to maximum height.If the total flight time is T, then the vertical velocity Vv isVv at t = 0,½*Vv at t = T/4,0 at t = T/2,-½Vv at t = 3T/4, and-Vv at t = T.Then the average velocity during the first interval is 3Vv/4, and the average velocity during the second interval is Vv/4.Then the vertical distance traveled during the first interval is 3 times the vertical distance during the second interval.Since the vertical distance traveled during the first interval was 3 m, then the vertical distance traveled during the second interval was 1 m, and the maximum height is 4 m.

range = (V²sin(2Θ))/g → 24m = V²sin(2Θ) / gmax height = (V·sinΘ)² / (2g) → 4m = V²sin²Θ / 2gDivide range by max height:24m / 4m = 6 = 2sin(2Θ) / sin²ΘUse trig identity sin(2Θ) = 2sinΘcosΘ:6 = 4sinΘcosΘ / sin²Θ = 4cosΘ / sinΘ = 4 / tanΘtanΘ = 4/6 = 2/3Θ = tan^-1(2/3) = assuming the ball just clears the wall.