Answer: [tex]\sqrt{2}\frac{V_{o}}{g}[/tex]
Explanation:
This situation is related to projectile motion and the main equation in this case is:
[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex]
Where:
[tex]y_{o}=0[/tex] is the initial height of the cannon ball
[tex]y=0[/tex] is the final height of the cannon ball
[tex]V_{o}[/tex] is the initial speed of the cannon ball
[tex]\theta=45\°[/tex] is the angle at which the cannon ball was fired
[tex]g[/tex] is the acceleration due gravity
[tex]t[/tex] is the time the cannon ball is in air
[tex]0=0+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex]
Isolating [tex]t[/tex]:
[tex]t=\frac{2V_{o}sin\theta}{g}[/tex]
[tex]t=\frac{2V_{o}sin(45\°)}{g}[/tex]
Since [tex]sin(45\°)=\frac{\sqrt{2}}{2}[/tex]:
[tex]t=2\frac{V_{o}}{g}\frac{\sqrt{2}}{2}[/tex]
Finally:
[tex]t=\sqrt{2}\frac{V_{o}}{g}[/tex]
