Answer:
[tex]y_m_a_x=10m[/tex]
Step-by-step explanation:
The max height of an object can be calculated using one the basic projectile motion equations:
[tex]y_m_a_x= \frac{v^2*sin^2(\theta)}{2g}[/tex]
Where:
[tex]g=Gravitational\hspace{3}constant\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}object\\\theta=Launch\hspace{3}angle[/tex]
In this case, let's asumme g=9.8m/s^2, besides θ=90, because the water ballon was threw straight up, so:
[tex]y_m_a_x= \frac{v^2*sin^2(\theta)}{2g}=\frac{(14^{2})*sin^2(90) }{2*9.8}=\frac{196*1}{19.6}=10m[/tex]
