Answer:
The difference is height is [tex]\Delta h =6.92 \ m[/tex]
Explanation:
From the question we are told that
   The distance of ball  from the goal is [tex]d = 36.0 \ m[/tex]
  The height of the crossbar is  [tex]h = 3.05 \ m[/tex]
    The speed of the ball is [tex]v = 21.6 \ m/s[/tex]
    The angle at which the ball was kicked is [tex]\theta = 50 ^o[/tex]
The height attained by the ball is mathematically represented as
   [tex]H = v_v * t - \frac{1}{2} gt^2[/tex]
Where [tex]v_v[/tex] is the vertical component of  velocity which is mathematically represented as
   [tex]v_v = v * sin (\theta )[/tex]
substituting values
   [tex]v_v = 21.6 * (sin (50 ))[/tex]
   [tex]v_v = 16.55 \ m/s[/tex]
Now the time taken is  evaluated as
    [tex]t = \frac{d}{v * cos(\theta )}[/tex]
substituting value
   [tex]t = \frac{36}{21.6 * cos(50 )}[/tex]
  [tex]t = 2.593 \ s[/tex]
So
   [tex]H = 16.55 * 2.593 - \frac{1}{2} * 9.8 * (2.593)^3[/tex]
   [tex]H = 9.97 \ m[/tex]
The difference  in height is mathematically evaluated as
   [tex]\Delta h = H - h[/tex]
substituting value
  [tex]\Delta h = 9.97 - 3.05[/tex]
  [tex]\Delta h =6.92 \ m[/tex]
