To solve this problem, it is necessary to apply the concepts related to the kinematic equations of movement description, both for the description of the position and of the acceleration and velocity.
The effective component that helps us find the time it takes to make the horizontal path is determined by the cosine, that is,
[tex]v = \frac{x}{t} \rightarrow t = v*x[/tex]
Where,
x = Displacement
t = time
[tex]t = \frac{17.5}{23.5cos(20.5)}[/tex]
t = 0.795s
With the net time found the maximum height would be described by the vertical component of the speed, therefore:
[tex]y = y_0+v_0t+\frac{1}{2}gt^2[/tex]
[tex]y = v_0sin(\theta)*t+\frac{1}{2}gt^2[/tex]
[tex]y = 23.5sin(20.5)(0.795)+\frac{1}{2}(9.8)(0.795)^2[/tex]
[tex]y = 9.64m[/tex]
Therefore the height would be 9.64m
