Time taken by the water balloon to reach the bottom will be given as
[tex]h = \frac{1}{2} gt^2[/tex]
here we know that
[tex]h = 18 m[/tex]
[tex]g = 9.8 m/s^2[/tex]
now by the above formula
[tex]18 = \frac{1}{2}*9.8* t^2[/tex]
[tex]18 = 4.9 t^2[/tex]
[tex]t = 1.92 s[/tex]
now in the same time interval we can say the distance moved by it will be
[tex]d = v_x * t[/tex]
[tex]d = 8.2 * 1.92 = 15.7 m[/tex]
so it will fall at a distance 15.7 m from its initial position
