Answer:
[tex]v_{y}=35.21m/s[/tex]
Explanation:
From the exercise we know the cannonball's initial velocity, the angle which its released with respect to the horizontal and its initial height
[tex]v_{o}=51.6m/s\\\beta =37.0º\\y_{o}=7m[/tex]
If we want to know whats the y-component of velocity we need to use the following formula:
[tex]v_{y}^2=v_{oy}^2+ag(y-y_{o})[/tex]
Knowing that [tex]g=-9.8m/s^2[/tex]
[tex]v_{y}=\sqrt{((51.6m/s)sin(37))^2-2(9.8m/s^2)(0m-7m)}=35.21m/s[/tex]
So, the cannonball's y-component of velocity is [tex]v_{y}=35.21m/s[/tex]
